洛谷 派对灯 Party Lamps

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USACO原题，比较恶心……看得题解
表示DFS只有三十分，就不用尝试DFS了，去看洛谷题解吧
题号：1468
https://www.luogu.org/wiki/show?name=%E9%A2%98%E8%A7%A3+P1468
作者: ☜闪耀星空☞ 更新时间: 2017-08-10 10:11

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#include<cstdlib>
#include<cstring>
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
int n, c, a, t;
bool light[10], dark[10],flag;
int map[10][10] = {
    { 0, 1, 1, 1, 1, 1, 1 }, //0
    { 0, 0, 0, 0, 0, 0, 0 }, //1
    { 0, 0, 1, 0, 1, 0, 1 }, //2
    { 0, 1, 0, 1, 0, 1, 0 }, //3
    { 0, 0, 1, 1, 0, 1, 1 }, //4
    { 0, 1, 0, 0, 1, 0, 0 }, //1,4
    { 0, 1, 1, 0, 0, 0, 1 }, //2,4
    { 0, 0, 0, 1, 1, 1, 0 }, //3,4
};
void print(int x){
    for (int i = 1; i <= 6; i++){
        if (light[i] && !map[x][i] || dark[i] && map[x][i])return;
    }
    flag = true;
    for (int i = 0; i < n; i++)cout << map[x][i % 6 + 1];
    cout << endl;
}
int main(){
    cin >> n >> c;
    while (cin >> t, t != -1)light[(t - 1) % 6 + 1] = true;
    while (cin >> t, t != -1)dark[(t - 1) % 6 + 1] = true;
    c = min(3, c);
    switch (c){
    case 0:print(0); break;
    case 1:print(1); print(2); print(4); print(3); break;
    case 2:print(1); print(7); print(2); print(4); print(3); print(6); print(0); break;
    case 3:print(1); print(7); print(2); print(4); print(5); print(3); print(6); print(0); break;
    }
    if (!flag)cout << "IMPOSSIBLE" << endl;
}

USACO Section 2.1 PROB The Castle

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The Castle
IOI’94 – Day 1
In a stroke of luck almost beyond imagination, Farmer John was sent a ticket to the Irish Sweepstakes (really a lottery) for his birthday. This ticket turned out to have only the winning number for the lottery! Farmer John won a fabulous castle in the Irish countryside.

Bragging rights being what they are in Wisconsin, Farmer John wished to tell his cows all about the castle. He wanted to know how many rooms it has and how big the largest room was. In fact, he wants to take out a single wall to make an even bigger room.

Your task is to help Farmer John know the exact room count and sizes.

The castle floorplan is divided into M (wide) by N (1 <=M,N<=50) square modules. Each such module can have between zero and four walls. Castles always have walls on their “outer edges” to keep out the wind and rain.

Consider this annotated floorplan of a castle:

     1   2   3   4   5   6   7
   #############################
 1 #   |   #   |   #   |   |   #
   #####---#####---#---#####---#   
 2 #   #   |   #   #   #   #   #
   #---#####---#####---#####---#
 3 #   |   |   #   #   #   #   #   
   #---#########---#####---#---#
 4 # ->#   |   |   |   |   #   #   
   ############################# 

#  = Wall     -,|  = No wall
-> = Points to the wall to remove to
     make the largest possible new room

By way of example, this castle sits on a 7 x 4 base. A “room” includes any set of connected “squares” in the floor plan. This floorplan contains five rooms (whose sizes are 9, 7, 3, 1, and 8 in no particular order).

Removing the wall marked by the arrow merges a pair of rooms to make the largest possible room that can be made by removing a single wall.

The castle always has at least two rooms and always has a wall that can be removed.

PROGRAM NAME: castle

INPUT FORMAT

The map is stored in the form of numbers, one number for each module, M numbers on each of N lines to describe the floorplan. The input order corresponds to the numbering in the example diagram above.

Each module number tells how many of the four walls exist and is the sum of up to four integers:

  • 1: wall to the west
  • 2: wall to the north
  • 4: wall to the east
  • 8: wall to the south

Inner walls are defined twice; a wall to the south in module 1,1 is also indicated as a wall to the north in module 2,1.

Line 1: Two space-separated integers: M and N
Line 2..: M x N integers, several per line.

SAMPLE INPUT (file castle.in)

7 4
11 6 11 6 3 10 6
7 9 6 13 5 15 5
1 10 12 7 13 7 5
13 11 10 8 10 12 13

OUTPUT FORMAT

The output contains several lines:

Line 1: The number of rooms the castle has.
Line 2: The size of the largest room
Line 3: The size of the largest room creatable by removing one wall
Line 4: The single wall to remove to make the largest room possible

Choose the optimal wall to remove from the set of optimal walls by choosing the module farthest to the west (and then, if still tied, farthest to the south). If still tied, choose ‘N’ before ‘E’. Name that wall by naming the module that borders it on either the west or south, along with a direction of N or E giving the location of the wall with respect to the module.

SAMPLE OUTPUT (file castle.out)

5
9
16
4 1 E

题解：无USACO输入输出，使用标准输入输出，题目可参见洛谷P1457

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#include<iostream>
#include<utility>
using namespace std;
int m, n;
int arr[55][55];
bool visited[55][55];
pair<int,int> father[55][55];
int mapsize[55][55];
inline int _find(int x, int y) {
    return mapsize[father[x][y].first][father[x][y].second];
}
inline int getBit(int n, int pos) {
    return (n >> pos) & 1;
}
int dfs(int x, int y,int fatherx,int fathery) {
    if (x<1||x>n||y<1||y>m||visited[x][y])return 0;
    visited[x][y] = true;
    father[x][y] = pair<int, int>(fatherx, fathery);
    int tot = 0;
    if (!getBit(arr[x][y], 0))tot += dfs(x , y-1, fatherx, fathery);
    if (!getBit(arr[x][y], 1))tot += dfs(x-1, y, fatherx, fathery);
    if (!getBit(arr[x][y], 2))tot += dfs(x, y+1, fatherx, fathery);
    if (!getBit(arr[x][y], 3))tot += dfs(x+1, y, fatherx, fathery);
    return tot+1;
}
int main() {
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            father[i][j] = pair<int, int>(-1, -1);
        }
    }
    cin >> m >> n;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> arr[i][j];
        }
    }
    int maxs = 0,room=0;

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            int t = dfs(i, j, i, j);
            if (t != 0) {
                room++;
                mapsize[i][j] = t;
            }
            if (t > maxs)maxs = t;
        }
    }
    cout << room << endl << maxs << endl;
    int maxcombsquare = 0, wallx=-1, wally=-1;
    char pos;
    for (int j = m; j >=1; j--){
        for (int i = 1; i<=n ; i++) {
            if (j!=m&&getBit(arr[i][j], 2) && father[i][j] != father[i][j + 1]) {
                if (_find(i, j) + _find(i, j + 1) >= maxcombsquare) {
                    maxcombsquare = _find(i, j) + _find(i, j + 1);
                    wallx = i; wally = j; pos = 'E';
                }
            }
            if (i != 1 && getBit(arr[i][j], 1) && father[i][j] != father[i - 1][j]) {
                if (_find(i, j) + _find(i - 1, j) >= maxcombsquare) {
                    maxcombsquare = _find(i, j) + _find(i - 1, j);
                    wallx = i; wally = j; pos = 'N';
                }
            }
        }
    }
    cout << maxcombsquare << endl << wallx << " " << wally << " " << pos << endl;
}

USACO Section 1.2 Transformations

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Transformations
A square pattern of size N x N (1 <= N <= 10) black and white square tiles is transformed into another square pattern. Write a program that will recognize the minimum transformation that has been applied to the original pattern given the following list of possible transformations: #1: 90 Degree Rotation: The pattern was rotated clockwise 90 degrees. #2: 180 Degree Rotation: The pattern was rotated clockwise 180 degrees. #3: 270 Degree Rotation: The pattern was rotated clockwise 270 degrees. #4: Reflection: The pattern was reflected horizontally (turned into a mirror image of itself by reflecting around a vertical line in the middle of the image). #5: Combination: The pattern was reflected horizontally and then subjected to one of the rotations (#1-#3). #6: No Change: The original pattern was not changed. #7: Invalid Transformation: The new pattern was not obtained by any of the above methods. In the case that more than one transform could have been used, choose the one with the minimum number above. PROGRAM NAME: transform INPUT FORMAT Line 1: A single integer, N Line 2..N+1: N lines of N characters (each either `@' or `-'); this is the square before transformation Line N+2..2*N+1: N lines of N characters (each either `@' or `-'); this is the square after transformation SAMPLE INPUT (file transform.in) 3 @-@ --- @@- @-@ @-- --@ OUTPUT FORMAT A single line containing the number from 1 through 7 (described above) that categorizes the transformation required to change from the `before' representation to the `after' representation. SAMPLE OUTPUT (file transform.out) 1 标准输入输出:

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#include<iostream>
#include<string>
#include<cstring>
int n;
namespace function {
    void reverse_str(char * s) {
        int len = strlen(s);
        for (int i = 0,j=len-1; i <j; i++,j--) {
            std::swap(s[i], s[j]);
        }
    }
    bool cmp(char arr1[11][11], char arr2[11][11]) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (arr1[i][j] != arr2[i][j])return false;
            }
        }
        return true;
    }
    void rotate90(char arr1[11][11], char arr2[11][11]) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                arr2[j][n-1-i] = arr1[i][j];
            }
        }
    }
    void rotate180(char arr1[11][11], char arr2[11][11]) {
        rotate90(arr1, arr2);
        rotate90(arr2, arr1);
        std::memcpy(arr2, arr1, sizeof(char)*11*11);
    }
    void rotate270(char arr1[11][11], char arr2[11][11]) {
        rotate90(arr1, arr2);
        rotate90(arr2, arr1);
        rotate90(arr1, arr2);
    }
    void reflection(char arr1[11][11], char arr2[11][11]) {
        std::memcpy(arr2, arr1, sizeof(char) * 11 * 11);
        for (int i = 0; i < n; i++) {
            reverse_str(arr2[i]);
        }
    }
    bool combination(char arr1[11][11], char arr2[11][11],char original[11][11],char transfered[11][11]) {
        std::memcpy(arr1, original, sizeof(char) * 11 * 11);
        reflection(arr1, arr2);
        std::memcpy(arr1, arr2, sizeof(char) * 11 * 11);
        rotate90(arr1, arr2);
        if (cmp(transfered, arr2))return true;
        rotate180(arr1, arr2);
        if (cmp(transfered, arr2))return true;
        std::memcpy(arr1, original, sizeof(char) * 11 * 11);
        reflection(arr1, arr2);
        std::memcpy(arr1, arr2, sizeof(char) * 11 * 11);
        rotate270(arr1, arr2);
        if (cmp(transfered, arr2))return true;
        return false;
    }
}
namespace std {
    char arr_original[11][11],arr_transfered[11][11];
    char temp1[11][11], temp2[11][11];
    int main() {
        function::rotate90(arr_original, temp1);
        if (function::cmp(temp1, arr_transfered))return 1;
        function::rotate90(temp1, temp2);
        if (function::cmp(temp2, arr_transfered))return 2;
        function::rotate90(temp2, temp1);
        if (function::cmp(temp1, arr_transfered))return 3;
        function::reflection(arr_original, temp1);
        if (function::cmp(temp1, arr_transfered))return 4;
        if(function::combination(temp1, temp2, arr_original,arr_transfered))return 5;
        if (function::cmp(arr_original, arr_transfered))return 6;
        return 7;
    }
}
int main() {
    std::ios::sync_with_stdio(false);
    std::cin >> n;
    for (int i = 0; i < n; i++) {
        std::cin >> std::arr_original[i];
    }
    for (int i = 0; i < n; i++) {
        std::cin >> std::arr_transfered[i];
    }
    std::cout<<std::main();
}

洛谷 幻想迷宫

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题号：1363
首先放上来一个AC代码。

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#include<iostream>
#include<cstring>
using namespace std;
int n, m;
char map[1505][1505];
int visited[1505][1505][2];
bool dfs(int orix, int oriy){
    int x = (orix%n+n) % n, y = (oriy%m+m) % m;
    if (map[x][y] != 'S' && map[x][y] != '.')return false;
    if (orix== visited[x][y][0]&&oriy==visited[x][y][1])return false;
    if (visited[x][y][0] != -1)return true;
    visited[x][y][0] = orix;
    visited[x][y][1] = oriy;
    if (dfs(orix + 1, oriy))return true;
    if (dfs(orix - 1, oriy))return true;
    if (dfs(orix, oriy + 1))return true;
    if (dfs(orix, oriy - 1))return true;
    return false;
}
int main(){
    ios::sync_with_stdio(false);
    while (cin >> n >> m){
        memset(visited, -1, sizeof(visited));
        for (int i = 0; i < n; i++)cin >> map[i];
        int sx, sy;
        for (int i = 0; i < n; i++){
            for (int j = 0; j < m; j++){
                if (map[i][j] == 'S'){
                    sx = i;
                    sy = j;
                    break;
                }
            }
        }
        bool flag = dfs(sx, sy);
        if (flag)cout << "Yes" << endl;
        else cout << "No" << endl;
    }
}

题解页面：https://www.luogu.org/wiki/show?name=%E9%A2%98%E8%A7%A3+P1363
该思路参考：作者: mike_he 更新时间: 2016-07-27 20:57
有两种思路是错误的，在下面列举一下：
继续阅读 →

洛谷 [USACO1.1]坏掉的项链Broken Necklace

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题号：1203
并不知道为什么AC，本来45分瞎改了改AC的。。。这破题也是神烦……

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#include<iostream>
#include<algorithm>
#include<string>
#include<utility>
#include<queue>
#include<cstdlib>
#include<cstring>
#include<map>
using namespace std;
int n;
string s;
char search(int start, int ptr){
    for (int i = ptr; i - start +1 <= n; i++){
        if (s[i] != 'w')return s[i];
    }
    return 'w';
}
int run(int start,char c, char a){
    int tmax = 0;
    bool flag = false;
    int ptr = start;
    char cur;
    if (s[start] == a)return 0;
    while (1){
        if (!flag){
            if (s[ptr] == c || s[ptr] == 'w')tmax++;
            else {
                flag = true;
                tmax++;
                cur = search(start, ptr); //就这里，我觉得reference 2应该是ptr+1才对。。
            }
        }
        else{
            if (s[ptr] == cur || s[ptr] == 'w')tmax++;
            else break;
        }
        ptr++;
        if (ptr - start >= n){
           
            break;
        }
    }
    return tmax;
}
int main(){
    ios::sync_with_stdio(false);
    cin >> n >> s;
    s.append(s);
    int m = 0;
    for (int i = 0; i <= n; i++){
        int t = run(i, 'b', 'r');
        if (t>m)m = t;
        t = run(i, 'r', 'b');
        if (t > m)m = t;
    }
    cout << m;
}

洛谷 过河

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题号：1052
参考题解：
https://www.luogu.org/wiki/show?name=%E9%A2%98%E8%A7%A3+P1052
作者: CoolTeam 更新时间: 2015-08-11 18:26
作者: shutdown 更新时间: 2013-07-30 20:20
作者: VengefulSpirit 更新时间: 2013-06-25 11:17
作者: sb123 更新时间: 2017-08-07 17:58

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#include<iostream>
#include<algorithm>
#include<string>
#include<utility>
#include<queue>
#include<cstdlib>
#include<cstring>
#include<map>
using namespace std;
int f[20005];
int Mk[105],M[105];
//ifstream cin("in.txt");
//ofstream cout("out.txt");
int main(){
    ios::sync_with_stdio(false);
    int l, s, t, m,ans=0;
    cin >> l >> s >> t >> m;
    for (int i = 1; i <= m; i++){
        cin >> Mk[i];//存入M的临时数组
    }
    if (s == t){//当s==t时是特殊情况，只需判断为s/t倍数的点有无石子即可 ；参考CoolTeam。为啥用动态规划过不了，我也不知道
        for (int i = 1; i <= m; i++){
            if (Mk[i] % s == 0)ans++;
        }
        cout << ans;
        return 0;
    } //否则需要使用动态规划
    memset(f, 0x7f, sizeof(f));
    f[0] = 0;
    sort(Mk + 1, Mk + m + 1); //先给临时数组排个序
    for (int i = 1; i <= m; i++){
        if (Mk[i] - Mk[i - 1] >= 100)M[i] = M[i - 1] + 100; //如果两相邻石子之间距离大于100，就将他们之间的距离缩小到100 ；参考shutdown的结论，VengefulSpirit的证明过程
        else M[i] = M[i - 1] + Mk[i] - Mk[i - 1];
    }
    int M_cnt = 1;
    for (int i = 1; i <= M[m]+t; i++){ //比较容易的动态规划 ；参考sb123的动规转移方程
        for (int j = i-s; j >= max(i - t,0); j--){
            f[i] = min(f[j], f[i]);
        }
        if (M[M_cnt] == i){
            f[i]++;
            M_cnt++;
        }
    }
    ans = 10000;
    for (int i = M[m] + t-1; i >= M[m]; i--)ans = min(ans, f[i]);  //遍历从最后一个石子到最后一个石子+t-1的最短距离
    cout << ans;
}

20180805更新，表示缩点/离散化恶心死了，写了半天60，迫不得已看题解了。、
参考题解：https://www.luogu.org/blog/GXu/solution-p1052
https://peerless.blog.luogu.org/guo-he-xie-ti-bao-gao

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#include<iostream>
#include<string>
#include<algorithm>
#include<functional>
#define ll long long
#define pii pair<int,int>
#define PINF 0x7fffffff
#define NINF 0x80000000
using namespace std;
int l;
int s, t, m;
int pos[105];
int d[105];
bool stone[400000];
int f[400000];
int main() {
    cin >> l >> s >> t >> m;
    for (int i = 1; i <= m; i++)cin >> pos[i];
    sort(pos + 1, pos + m + 1);
    for (int i = 1; i <= m; i++)d[i] = (pos[i] - pos[i - 1]) % 2520;
    for (int i = 1; i <= m; i++) {
        pos[i] = pos[i - 1] + d[i];
        stone[pos[i]] = true;
    }
    l = pos[m];
    for (int i = 1; i <= l + t; i++)f[i] = m;
    for (int i = 1; i <= l + t; i++) {
        for (int j = s; j <= t; j++) {
            if (i >= j)f[i] = min(f[i], f[i - j]);
            f[i] += stone[i];
        }
    }
    int ans = m;
    for (int i = l; i < l + t; i++)ans = min(ans, f[i]);
    cout << ans;
}

洛谷 A % B Problem

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题目编号：1865
用到的思想，一维前缀和，埃式素数筛

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#include<string>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<iostream>
using namespace std;
bool prime[1000005];
int prefixsum[1000005];
int main(){
    ios::sync_with_stdio(false);
    int n,m;
    cin>>n>>m;
    memset(prime,true,sizeof(prime));
    prime[1]=false;
    for(int i=2;i<=m;i++){
        if(!prime[i])continue;
        for(int j=2;i*j<=m;j++){
            prime[i*j]=false;
        }
    }
    for(int i=1;i<=m;i++){
        prefixsum[i]=prefixsum[i-1]+prime[i];
    }
    for(int i=1;i<=n;i++){
        int l,r;
        cin>>l>>r;
        if(l<1||r>m){
            cout<<"Crossing the line"<<endl;
            continue;
        }
        cout<<prefixsum[r]-prefixsum[l-1]<<endl;
    }
}

洛谷 最大正方形

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题号：1387
参考题解：https://www.luogu.org/wiki/show?name=%E9%A2%98%E8%A7%A3+P1387
作者: hychychyc 更新时间: 2017-06-06 11:25

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#include<iostream>
#include<algorithm>
#include<string>
#include<utility>
#include<queue>
#include<cstdlib>
#include<cstring>
#include<map>
using namespace std;
int n, m;
int arr[105][105];
int sum[105][105];
int main(){
    ios::sync_with_stdio(false);
    cin >> n >> m;
    for (int i = 1; i <= n; i++){
        for (int j = 1; j <= m; j++){
            cin >> arr[i][j];
        }
    }
    for (int i = 1; i <= n; i++){
        for (int j = 1; j <= m; j++){
            sum[i][j] = sum[i][j - 1] + sum[i - 1][j] - sum[i - 1][j - 1] + arr[i][j]; //二维前缀和
        }
    }
    int maxn = 1;
    for (int i = 1; i <= n; i++){
        for (int j = 1; j <= m; j++){
            for (int e = maxn; i + e <= n&&j + e <= m; e++){
                int w = sum[i + e][j + e] - sum[i][j + e] - sum[i + e][j] + sum[i][j];  //这个用二维前缀和计算正方形和
                if (e*e == w){
                    maxn = max(maxn, e);
                }
                else
                    break;
            }
        }
    }
    cout << maxn;
}

放上一张前缀和的图片，写的也很不清楚，也许有助于理解。

20180802更新：
上面代码都是错误的！！！！！！！！！
二维前缀和方程是：

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#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
int n, m;
int arr[105][105];
int main(){
    cin >> n >> m;
    for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++)cin >> arr[i][j];
    for (int i = 1; i <= n; i++)for (int j = 1; j <= m; j++)arr[i][j] = arr[i - 1][j] + arr[i][j - 1] - arr[i - 1][j - 1] + arr[i][j];
    for (int k = min(n, m); k > 0; k--) {
        for (int i = 1; i <= n - k + 1; i++) {
            for (int j = 1; j <= m - k + 1; j++) {
                if (k*k == arr[i + k - 1][j + k - 1] - arr[i - 1][j + k - 1] - arr[i + k - 1][j - 1] + arr[i - 1][j - 1]) { //参考这里
                    cout << k;
                    return 0;
                }
            }
        }
    }
}

动态规划做法：题解https://www.luogu.org/blog/user23035/solution-p1387

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#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
int n, m;
int arr[105][105];
int main() {
    cin >> n >> m;
    int ans = 0;
    for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) {
        cin >> arr[i][j];
        if(arr[i][j])arr[i][j] = min(min(arr[i - 1][j], arr[i][j - 1]), arr[i - 1][j - 1]) + 1;
        ans = max(ans, arr[i][j]);
    }
    cout << ans;
}

洛谷 传纸条

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题号：1006
同方格取数，只不过是有个别的地方需要一些变动。我傻傻的有一些地方没变……2333

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#include<iostream>
#include<algorithm>
#include<string>
#include<utility>
#include<queue>
#include<cstdlib>
#include<cstring>
#include<map>
using namespace std;
int m,n;
int f[55][55][55][55],arr[55][55];
int main(){
    ios::sync_with_stdio(false);
    cin >> m >> n;
    for (int i = 1; i <= m; i++){
        for (int j = 1; j <= n; j++){
            cin >> arr[i][j];
        }
    }
    for (int i = 1; i <= m; i++){
        for (int j = 1; j <= n; j++){
            for (int k = 1; k <= m; k++){
                for (int l = 1; l <= n; l++){
                    f[i][j][k][l] = max(f[i - 1][j][k - 1][l], f[i][j - 1][k - 1][l]);
                    f[i][j][k][l] = max(f[i][j][k][l], f[i - 1][j][k][l - 1]);
                    f[i][j][k][l] = max(f[i][j][k][l], f[i][j - 1][k][l - 1]);
                    if (i == k && j == l)f[i][j][k][l] += arr[i][j];
                    else f[i][j][k][l] += arr[i][j] + arr[k][l];
                }
            }
        }
    }
    cout << f[m][n][m][n];
}

20180802
做的烦得要死。。。上面的代码没看懂^^^^……………….
https://www.luogu.org/blog/user20197/solution-p1006
https://www.luogu.org/blog/user26182/solution-p1006

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#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
int m, n;
int arr[205][205];
int f[55][55][55][55];
int main() {
    cin >> m >> n;
    for (int i = 1; i <= m; i++)for (int j = 1; j <= n; j++)cin >> arr[i][j];
    for (int i = 1; i <= m; i++)for (int j = 1; j <= n; j++)for (int k = 1; k <= m; k++)for (int l = j + 1; l <= n; l++) {
        f[i][j][k][l] = max(max(f[i - 1][j][k - 1][l], f[i - 1][j][k][l - 1]), max(f[i][j - 1][k - 1][l], f[i][j - 1][k][l - 1])) + arr[i][j] + arr[k][l];
    }
    cout << f[m][n-1][m-1][n];
}

洛谷 方格取数

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题号：1004
这题还是比较简单的……虽然我看了题解，但是还是能看得懂的。。
参考题解：https://www.luogu.org/wiki/show?name=%E9%A2%98%E8%A7%A3+P1004
作者: sb123 更新时间: 2017-08-07 14:52

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#include<iostream>
#include<algorithm>
#include<string>
#include<utility>
#include<queue>
#include<cstdlib>
#include<cstring>
#include<map>
using namespace std;
int n;
int f[12][12][12][12],arr[12][12];
int main(){
    ios::sync_with_stdio(false);
    cin >> n;
    int rf1, rf2, rf3;
    while (cin >> rf1 >> rf2 >> rf3){
        if (rf1 == 0 && rf2 == 0 && rf3 == 0)break;
        arr[rf1][rf2] = rf3;
    }
    for (int i = 1; i <= n; i++){
        for (int j = 1; j <= n; j++){
            for (int k = 1; k <= n; k++){
                for (int l = 1; l <= n; l++){
                    f[i][j][k][l] = max(f[i - 1][j][k - 1][l], f[i][j - 1][k - 1][l]);
                    f[i][j][k][l] = max(f[i][j][k][l], f[i - 1][j][k][l - 1]);
                    f[i][j][k][l] = max(f[i][j][k][l], f[i][j - 1][k][l - 1]);
                    if (i == k && j == l)f[i][j][k][l] += arr[i][j];
                    else f[i][j][k][l] += arr[i][j] + arr[k][l];
                }
            }
        }
    }
    cout << f[n][n][n][n];
}