NOIP2017 普及组第三道 棋盘

这题写的异常难受……先想着用动态规划,难受死了,才20分。然后后来再看题解,用记忆化BFS,写的也难受,主要是太难想,终于AC了。开心~

写了有6、7个钟头吧。挺羞愧的。。。我写这种题怎么能这么长时间呢……

所有测试点全部0ms。

成就感极高。

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#include<iostream>
#include<memory.h>
#include<queue>
using namespace std;
struct qs { //Convenient for passing parameters
    int tx, ty, os, rs, sc; //destination's coordinate (x,y) :tx,ty ;previous point::original status of color,real status of color :os,rs ;previous point's score:sc.
    qs(int tx1, int ty1, int os1, int rs1, int sc1) {
        tx = tx1; ty = ty1; os = os1; rs = rs1; sc = sc1;
    } //the init of struct
};
int m, n;
int arr[105][105], f[105][105]; //arr stores the color(status) of all coordinate; f stores the least gold the person will spend
int shiftx[4] = { 1,-1,0,0 }, shifty[4] = { 0,0,1,-1 }; //shifts for the coordinate
queue<qs> q;
void execute(qs o) {
    //Calculate New Score
    //Compare Which one is smaller
    //Only when the new one is smaller one,will the program insert the queue
    int nscore = 0x7fffffff;
    if (o.rs != -1 && arr[o.tx][o.ty] != -1) { // C2C(R2R/Y2Y/R2Y/Y2R) {"C":"Color","2":"To","R":"Red","Y":"Yellow"};
        if (o.rs == arr[o.tx][o.ty])nscore = o.sc; // R2R/Y2Y
        else nscore = o.sc + 1; // R2Y/T2R
        if (nscore < f[o.tx][o.ty] || o.tx==1&&o.ty==1) { //o.tx==1&&o.ty==1 is to create an entrance for the inital queue push
            f[o.tx][o.ty] = nscore;
            for (int i = 0; i < 4; i++)q.push(qs(o.tx + shiftx[i], o.ty + shifty[i], arr[o.tx][o.ty], arr[o.tx][o.ty], nscore));
        }
    }
    else if (o.os != -1 && arr[o.tx][o.ty] == -1) { // C2N
        if (o.os == 0 && o.sc + 2 < f[o.tx][o.ty]) { //Assume that the temporary color of current cell is Red
            f[o.tx][o.ty] = nscore = o.sc + 2;
            for (int i = 0; i < 4; i++)q.push(qs(o.tx + shiftx[i], o.ty + shifty[i], -1, 0, nscore));
            nscore += 1;
            for (int i = 0; i < 4; i++)q.push(qs(o.tx + shiftx[i], o.ty + shifty[i], -1, 1, nscore));
        }
        else if (o.os == 1 && o.sc + 2 < f[o.tx][o.ty]) { //Assume that the temporary color of current cell is Yellow
            f[o.tx][o.ty] = nscore = o.sc + 2;
            for (int i = 0; i < 4; i++)q.push(qs(o.tx + shiftx[i], o.ty + shifty[i], -1, 1, nscore));
            nscore += 1;
            for (int i = 0; i < 4; i++)q.push(qs(o.tx + shiftx[i], o.ty + shifty[i], -1, 0, nscore));
        }
    }
}
int main() {
    memset(arr, -1, sizeof(arr));
    memset(f, 0x7f, sizeof(f));
    ios::sync_with_stdio(false);
    cin >> m >> n;
    for (int i = 1; i <= n; i++) {
        int x, y, c;
        cin >> x >> y >> c;
        arr[x][y] = c;
    }
    f[1][1] = 0;
    q.push(qs(1, 1, arr[1][1], arr[1][1], 0));
    while (!q.empty()) {
        qs cur = q.front();
        q.pop();
        if (cur.tx > m || cur.ty > m || cur.tx < 1 || cur.ty < 1)continue;
        execute(cur);
    }
    cout << (f[m][m] == 0x7f7f7f7f ? -1 : f[m][m]) << endl;
}

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