分类目录归档：图论

CSP 202009-3 点亮数字人生

拓扑排序，另外这道题字符串处理太多了，麻烦死了。这种第三题全都是大模拟。非常麻烦。

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#include <iostream>
#include <algorithm>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <string>
#include <unordered_map>
#include <vector>

using namespace std;

struct node {
string type;
int input;
string signals[10];
bool value;
};

void topoSortDegreeMinus(int id, vector<int> &inDegree, const vector<vector<int>> &edges) {
if (inDegree[id] != 0)return;
inDegree[id] = -1;
vector<int> storage;
for (int i = 0; i < edges[id].size(); i++) {
inDegree[edges[id][i]]--;
if (inDegree[edges[id][i]] == 0)
storage.push_back(i);
}
for (int i = 0; i < storage.size(); i++) {
topoSortDegreeMinus(storage[i], inDegree, edges);
}
}

bool checkCircle(int n, unordered_map<string, node> &nodeMap) { //Topological sort
vector<int> inDegree;
inDegree.resize(n+1);
vector<vector<int>> edges;
edges.resize(n+1);

for (int i = 1; i <= n; i++) {
string nodeID = "O" + to_string(i);
const node &curNode = nodeMap[nodeID];
for (int j = 0; j < curNode.input; j++) {
if (curNode.signals[j][0] == 'O') {
inDegree[i]++;
int edgeStart = strtol(&curNode.signals[j].c_str()[1], NULL, 10);
edges[edgeStart].push_back(i);
}
}
}

bool flag = false;
do {
flag = false;
for (int i = 1; i <= n; i++) {
if (inDegree[i] == 0) {
topoSortDegreeMinus(i, inDegree, edges);
flag = true;
}
}
} while (flag);

for (int i = 1; i <= n; i++) {
if (inDegree[i] > 0)
return true;
}
return false;
}

bool calculateLogicDFS(unordered_map<string, node> &nodeMap,
unordered_map<string, bool> &visited, string ID) {
node &curNode = nodeMap[ID];
if (visited[ID])return curNode.value;
visited[ID] = true;
if (ID[0] == 'I')return curNode.value;
if (curNode.type == "NOT") {
calculateLogicDFS(nodeMap, visited, curNode.signals[0]);
curNode.value=!nodeMap[curNode.signals[0]].value;
} else if (curNode.type == "AND") {
curNode.value = true;
for(int i=0;i<curNode.input;i++){
calculateLogicDFS(nodeMap, visited, curNode.signals[i]);
curNode.value &= nodeMap[curNode.signals[i]].value;
if(!curNode.value)break;
}
} else if (curNode.type == "OR") {
curNode.value = false;
for(int i=0;i<curNode.input;i++){
calculateLogicDFS(nodeMap, visited, curNode.signals[i]);
curNode.value |= nodeMap[curNode.signals[i]].value;
if(curNode.value)break;
}
} else if (curNode.type == "XOR") {
curNode.value = false;
for(int i=0;i<curNode.input;i++){
calculateLogicDFS(nodeMap, visited, curNode.signals[i]);
curNode.value ^= nodeMap[curNode.signals[i]].value;
}
} else if (curNode.type == "NAND") {
curNode.value = true;
for(int i=0;i<curNode.input;i++){
calculateLogicDFS(nodeMap, visited, curNode.signals[i]);
curNode.value &= nodeMap[curNode.signals[i]].value;
if(!curNode.value)break;
}
curNode.value=!curNode.value;
} else if (curNode.type == "NOR") {
curNode.value = false;
for(int i=0;i<curNode.input;i++){
calculateLogicDFS(nodeMap, visited, curNode.signals[i]);
curNode.value |= nodeMap[curNode.signals[i]].value;
if(curNode.value)break;
}
curNode.value=!curNode.value;
}
return curNode.value;
}

void calculateLogic(unordered_map<string, node> &nodeMap) {
unordered_map<string, bool> visited;
for(auto iter=nodeMap.begin();iter!=nodeMap.end();iter++){
if(!visited[iter->first])
calculateLogicDFS(nodeMap,visited,iter->first);
}
}

void solve() {
// Read Part 1
int m, n;
cin >> m >> n;
unordered_map<string, node> nodeMap;
for (int i = 1; i <= n; i++) {
node curNode;
cin >> curNode.type >> curNode.input;
for (int j = 0; j < curNode.input; j++) {
cin >> curNode.signals[j];
}
nodeMap["O" + to_string(i)] = curNode;
}

// Check Circle: Topological Sort
bool haveCircle = checkCircle(n, nodeMap);

// Read Part 2
int s;
cin >> s;
vector<vector<int>> inputs;
inputs.resize(s);
for (int i = 0; i < s; i++) {
inputs[i].push_back(m);
int t;
for (int j = 0; j < m; j++) {
cin >> t;
inputs[i].push_back(t);
}
}
vector<vector<int>> outputs;
outputs.resize(s);
for (int i = 0; i < s; i++) {
int k;
cin >> k;
outputs[i].push_back(k);
int t;
for (int j = 1; j <= k; j++) {
cin >> t;
outputs[i].push_back(t);
}
}

if (haveCircle) {
cout << "LOOP\n";
return;
}
for (int i = 0; i < s; i++) {
for (int j = 1; j <= m; j++) {
node tmpNode;
tmpNode.value = inputs[i][j];
nodeMap["I" + to_string(j)] = tmpNode;
}
calculateLogic(nodeMap);
for(int j=1;j<=outputs[i][0];j++){
cout<<nodeMap["O"+to_string(outputs[i][j])].value<<" ";
}
cout<<"\n";
}
}

int main() {
ios::sync_with_stdio(false);
int q;
cin >> q;
while (q--)
solve();
}

Openjudge BJUTACM 19g系列代码

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题目链接：
http://bjutacm.openjudge.cn/lianxi/?page=2

继续阅读 →

洛谷 P3381 【模板】最小费用最大流

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参考链接：https://www.cnblogs.com/widerg/p/9394929.html

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#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;
const int eN = 100005;
const int vN = 10005;
const int INF = 0x7f7f7f7f;
int n, m, s, t;
int ePtr = -1; //e的实际存储只能从偶数号顶点开始，奇数号顶点存储反向边
int to[eN << 1], nxt[eN << 1], value[eN << 1], cost[eN << 1];
int head[vN], dis[vN], minV[vN];
int preID[eN << 1], preNode[eN << 1];
bool inqueue[vN];
inline void createEdge(int u, int v, int w, int c) {
to[++ePtr] = v;
value[ePtr] = w;
cost[ePtr] = c;
nxt[ePtr] = head[u];
head[u] = ePtr;
}
bool SPFA(int s, int t) {
queue<int> q;
memset(dis, 0x7f, sizeof(dis));
memset(inqueue, false, sizeof(inqueue));
memset(preID, -1, sizeof(preID));
memset(preNode, -1, sizeof(preNode));
memset(minV, 0x7f, sizeof(minV));
dis[s] = 0;
inqueue[s] = true;
q.push(s);
while (!q.empty()) {
int x = q.front();
q.pop();
inqueue[x] = false;
for (int i = head[x]; ~i; i = nxt[i]) {
int dest = to[i];
if (dis[dest] > dis[x] + cost[i] && value[i]) {
dis[dest] = dis[x] + cost[i];
minV[dest] = min(minV[x], value[i]);
preID[dest] = i;
preNode[dest] = x;
if (!inqueue[dest]) {
inqueue[dest] = true;
q.push(dest);
}
}
}
}
return dis[t] != INF;
}
void MinCostMaxFlow(int s, int t, int& maxflow, int& mincost) {
while (SPFA(s, t)) {
for (int i = t; i != s; i = preNode[i]) {
value[preID[i]] -= minV[t];
value[preID[i] ^ 1] += minV[t];
}
maxflow += minV[t];
mincost += minV[t] * dis[t];
}
}
int main() {
memset(head, -1, sizeof(head));
scanf("%d%d%d%d", &n, &m, &s, &t);
for (int i = 1; i <= m; i++) {
int u, v, w, f;
scanf("%d%d%d%d", &u, &v, &w, &f);
createEdge(u, v, w, f);
createEdge(v, u, 0, -f);//第0号边被占用，这条语句为反向边留下空间
}
int maxflow = 0, mincost = 0;
MinCostMaxFlow(s, t, maxflow, mincost);
printf("%d %d\n", maxflow, mincost);
}

洛谷 P3376 【模板】网络最大流

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参考链接：https://www.cnblogs.com/widerg/p/9387909.html

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#include<iostream>
#include<algorithm>
#include<queue>
#include<cstring>
using namespace std;
const int eN = 100005;
const int vN = 10005;
const int INF = 0x7f7f7f7f;
int n, m, s, t;
int ePtr = -1; //e的实际存储只能从偶数号顶点开始，奇数号顶点存储反向边
int to[eN<<1], nxt[eN<<1], value[eN<<1];
int head[vN], dis[vN];
inline void createEdge(int u, int v, int w) {
to[++ePtr] = v;
value[ePtr] = w;
nxt[ePtr] = head[u];
head[u] = ePtr;
}
bool bfs(int s, int t) {
queue<int> q;
memset(dis, 0x7f, sizeof(dis));
dis[s] = 0;
q.push(s);
while (!q.empty()) {
int x = q.front();
q.pop();
for (int i = head[x]; ~i; i = nxt[i]) {
int dest = to[i];
if (dis[dest] == INF && value[i] != 0) {
dis[dest] = dis[x] + 1;
q.push(dest);
}
}
}
return dis[t] != INF;
}
int dfs(int x, int t, int maxflow) {
if (x == t)return maxflow;
int ans = 0;
for (int i = head[x]; ~i; i = nxt[i]) {
int dest = to[i];
if (dis[dest] != dis[x] + 1 || value[i] == 0 || ans >= maxflow)continue;
int f = dfs(dest, t, min(value[i], maxflow - ans));
value[i] -= f;
value[i ^ 1] += f;
ans += f;
}
return ans;
}
int dinic(int s, int t) {
int ans = 0;
while (bfs(s, t))ans += dfs(s, t, INF);
return ans;
}
int main() {
memset(head, -1, sizeof(head));
scanf("%d%d%d%d", &n, &m, &s, &t);
for (int i = 1; i <= m; i++) {
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
createEdge(u, v, w);
createEdge(v, u, 0);//第0号边被占用，这条语句为下一个边留下空间
}
printf("%d", dinic(s, t));
}

洛谷 P3386 【模板】二分图匹配

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#include<iostream>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<vector>
#include<list>
#include<cmath>
#include<queue>
#include<unordered_set>
#include<unordered_map>
using namespace std;
int n, m, e;
vector<int> edges[2005];
bool visited[2005];
int match[2005];
bool biFind(int node) {
if (visited[node])return false;
visited[node] = true;
for (int i = 0; i < edges[node].size(); i++) {
if (!match[edges[node][i]] || biFind(match[edges[node][i]])) {
match[node] = edges[node][i];
match[edges[node][i]] = node;
return true;
}
}
return false;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
cin >> n >> m >> e;
for (int i = 1; i <= e; i++) {
int u, v;
cin >> u >> v;
if (u > n || v > m)continue;
edges[u].push_back(v + n);
edges[v + n].push_back(u);
}
int cnter = 0;
for (int i = 1; i <= n; i++) {
memset(visited, false, sizeof(visited));
if (biFind(i))cnter++;
}
cout << cnter;
}

计蒜客 Random Access Iterator

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原题链接：https://nanti.jisuanke.com/t/41392
竟然做出来一道这么复杂的概率+DFS+逆元题，开心到爆炸，特此发题解一篇^v^

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#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
#include<vector>
#define MOD 1000000007
#define ll long long
using namespace std;
int n;
int MDep;
vector<int> edges[1000005];
bool visited[1000005];
ll quickpow(ll a, ll b) {
if (b == 0)return 1;
ll re = quickpow(a, b / 2) % MOD;
re = (re * re) % MOD;
if (b % 2 == 1)re *= a % MOD;
return re % MOD;
}
inline ll inv(ll x) {
return quickpow(x, MOD - 2) % MOD;
}
inline ll chance1(ll TotNode) {
ll ch = (inv(TotNode)) % MOD;
return ch;
}
inline ll chance2(ll onceCh,ll TotNode) {
ll onceNonch = (1ll-onceCh+MOD) % MOD;
ll nonch = quickpow(onceNonch, TotNode);
ll fin = (1ll - nonch + MOD) % MOD;
return fin;
}
int dfsMaxDep(int node) {
int maxDep = 1;
bool hasChild = false;
visited[node] = true;
for (int i = 0; i < edges[node].size(); i++) {
if (visited[edges[node][i]])continue;
hasChild = true;
maxDep = max(maxDep, dfsMaxDep(edges[node][i]) + 1);
}
return maxDep;
}
ll dfsChance(int depth,int node){
if (depth == MDep)return 1;
ll ch = 0,sCh=chance1(node==1?edges[node].size(): edges[node].size()-1);
visited[node] = true;
for (int i = 0; i < edges[node].size(); i++) {
if (visited[edges[node][i]])continue;
ll CurCH = dfsChance(depth + 1, edges[node][i]);
if (CurCH) {
ch += (CurCH * sCh) % MOD;
ch %= MOD;
}
}
return chance2(ch, node == 1 ? edges[node].size() : edges[node].size() - 1);
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n;
for (int i = 1; i < n; i++) {
int u, v;
cin >> u >> v;
edges[u].push_back(v);
edges[v].push_back(u);
}
MDep = dfsMaxDep(1);
memset(visited, false, sizeof(visited));
ll ch = dfsChance(1, 1);
cout << ch;
}

洛谷 P1113 杂务

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这题本来就是个拓扑排序，搞了半天竟然搞不出来，想了半天。。
一开始的想法是：搞拓扑排序，一轮一轮的扫描，每一轮出一个最大时间，把最大时间相加就好了，后来发现全WA，为什么呢？因为同一轮里出来的任务其实有可能是不互相依赖的。比如任务1时间需要10；任务2时间需要5；任务3需要在任务1之后，任务4需要在任务2之后。如果一轮一轮的扫描，那么必须10秒之后同时开始做任务3/4。而实际情况则应该是任务2，5个时间作完之后就可以做任务4了。根据失误我又想着反向建图，用DFS递归解决，对每个结点（及其之前节点）找其最大的所用时间，写出来AC了（对于本题，杂务 k(k>1)的准备工作只可能在杂务 1 至 k-1 中，如果没有这个要求，反向建图就很好）。但是其实我想多了。直接一开始的拓扑排序，顺着推就可以了。
DFS版：

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#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int n,times[10005],inDegree[10005],f[10005];
vector<int> edges[10005];
vector<int> curTasks;
int dfs(int node){
if(f[node]!=0)return f[node];
if(edges[node].size()==0)return f[node]=times[node];
int maxT=0;
for(int i=0;i<edges[node].size();i++){
maxT=max(maxT,dfs(edges[node][i]));
}
return f[node]=maxT+times[node];
}
int main(){
ios::sync_with_stdio(false);
cin>>n;
for(int i=1;i<=n;i++){
int v;
cin>>v>>times[i];
while(1){
cin>>v;
if(v==0)break;
edges[i].push_back(v);
inDegree[v]++;
}
}
int maxTime=0;
for(int i=1;i<=n;i++){
if(inDegree[i]==0)maxTime=max(maxTime,dfs(i));
}
cout<<maxTime;
}

正常版：

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#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int n,times[10005],inDegree[10005],allTimes[10005];
vector<int> edges[10005];
int sumTime=0;
int main(){
ios::sync_with_stdio(false);
cin>>n;
for(int i=1;i<=n;i++){
int v;
cin>>v>>times[i];
while(1){
cin>>v;
if(v==0)break;
edges[v].push_back(i);
inDegree[i]++;
}
}
int maxT=0;
for(int i=1;i<=n;i++)if(inDegree[i]==0){
allTimes[i]+=times[i];
for(int j=0;j<edges[i].size();j++){
allTimes[edges[i][j]]=max(allTimes[edges[i][j]],allTimes[i]);
inDegree[edges[i][j]]--;
}
inDegree[i]--;
maxT=max(maxT,allTimes[i]);
}
cout<<maxT;
}

洛谷 P1144 最短路计数

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自己一开始写了个，才60分，效率不高，后来看了题解。
参考题解：https://www.luogu.org/blog/user43513/luo-gu-p1144-zui-duan-lu-ji-shuo-ti-xie
60分代码：

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#include<iostream>
#include<cstring>
#include<queue>
#define MAXN 1000000
#define MAXM 2000000
#define MOD 100003
using namespace std;
int n,m;
int v[MAXM+5],nxt[MAXM+5];
bool visited[MAXM+5];
int fst[MAXN+5];
int dis[MAXN+5];
int cnter[MAXN+5];
int edgePtr=0;
void edgePush(int s,int e){
edgePtr++;
nxt[edgePtr]=fst[s];
v[edgePtr]=e;
fst[s]=edgePtr;
}
void SP(){
memset(visited,false,sizeof(visited));
memset(dis,0x7f,sizeof(dis));
queue<int> q,qd;
q.push(1);qd.push(0);
while(!q.empty()){
int node=q.front(),dist=qd.front();
q.pop();qd.pop();
if(visited[node])continue;
visited[node]=true;
dis[node]=dist;
for(int e=fst[node];e!=-1;e=nxt[e]){
if(!visited[v[e]]&&dis[v[e]]>dis[node]+1){
dis[v[e]]=dis[node]+1;
q.push(v[e]);qd.push(dis[v[e]]);
}
}
}
}
void DFS(int node,int dist){
if(dist!=dis[node])return;
cnter[node]++;
cnter[node]%=MOD;
for(int e=fst[node];e!=-1;e=nxt[e]){
if(visited[e])continue;
visited[e]=true;
DFS(v[e],dist+1);
visited[e]=false;
}
}
int main(){
memset(fst,-1,sizeof(fst));
ios::sync_with_stdio(false);
cin>>n>>m;
for(int i=1;i<=m;i++){
int x,y;
cin>>x>>y;
if(x==y)continue;
edgePush(x,y);
edgePush(y,x);
}
SP();
memset(visited,false,sizeof(visited));
DFS(1,0);
for(int i=1;i<=n;i++)cout<<cnter[i]<<endl;
}

100分的：

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#include<iostream>
#include<cstring>
#include<queue>
#define MAXN 1000000
#define MAXM 2000000
#define MOD 100003
using namespace std;
int n,m;
int v[MAXM+5],nxt[MAXM+5];
bool visited[MAXM+5];
int fst[MAXN+5];
int dis[MAXN+5];
int cnter[MAXN+5];
int edgePtr=0;
void edgePush(int s,int e){
edgePtr++;
nxt[edgePtr]=fst[s];
v[edgePtr]=e;
fst[s]=edgePtr;
}
void SP(){
memset(visited,false,sizeof(visited));
memset(dis,0x7f,sizeof(dis));
queue<int> q,qd;
q.push(1);qd.push(0);
cnter[1]=1;
while(!q.empty()){
int node=q.front(),dist=qd.front();
q.pop();qd.pop();
if(visited[node])continue;
visited[node]=true;
dis[node]=dist;
for(int e=fst[node];e!=-1;e=nxt[e]){
if(!visited[v[e]]&&dis[v[e]]>dis[node]+1){
dis[v[e]]=dis[node]+1;
q.push(v[e]);qd.push(dis[v[e]]);
cnter[v[e]]+=cnter[node];
cnter[v[e]]%=MOD;
}else if(dis[v[e]]==dis[node]+1){
cnter[v[e]]+=cnter[node];
cnter[v[e]]%=MOD;
}
}
}
}
int main(){
memset(fst,-1,sizeof(fst));
ios::sync_with_stdio(false);
cin>>n>>m;
for(int i=1;i<=m;i++){
int x,y;
cin>>x>>y;
if(x==y)continue;
edgePush(x,y);
edgePush(y,x);
}
SP();
for(int i=1;i<=n;i++)cout<<cnter[i]<<endl;
}

洛谷 P1346 电车

发表评论

基本上是裸的最短路，是第一个不用扳道岔的边长就是0，后面的需要扳道岔的长度就是1。然后愉快的套一个Dijkstra板子就可以了。

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#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
using namespace std;
int n,a,b;
struct edge{
int v,w;
edge(int vv,int ww){v=vv;w=ww;}
bool operator < (const edge& e2) const{
return w>e2.w;
}
};
vector<edge> edges[105];
priority_queue<edge> pq;
bool visited[105];
int dis[105];
int main(){
ios::sync_with_stdio(false);
cin>>n>>a>>b;
for(int i=1;i<=n;i++){
int t;
cin>>t;
for(int j=1;j<=t;j++){
int v;
cin>>v;
if(j==1)edges[i].push_back(edge(v,0));
else edges[i].push_back(edge(v,1));
}
}
//Dijkstra
memset(dis,0x7f,sizeof(dis));
pq.push(edge(a,0));
while(!pq.empty()){
edge cur=pq.top();
pq.pop();
if(visited[cur.v])continue;
visited[cur.v]=true;
dis[cur.v]=cur.w;
for(int i=0;i<edges[cur.v].size();i++){
if(!visited[edges[cur.v][i].v]&&dis[edges[cur.v][i].v]>dis[cur.v]+edges[cur.v][i].w){
dis[edges[cur.v][i].v]=dis[cur.v]+edges[cur.v][i].w;
pq.push(edge(edges[cur.v][i].v,dis[edges[cur.v][i].v]));
}
}
}
if(dis[b]!=0x7f7f7f7f)cout<<dis[b];
else cout<<-1;
}

洛谷 P1991 无线通讯网

发表评论

这道题很显然能够看出，只需找出与卫星电话数相等的联通块数，然后取联通块里的最大边就可以了。但是我想不出来怎么好写。
看过题解之后，经题解提醒“树的性质：删掉n条边一定出现n+1个连通块”，这样就好做了。

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#include<iostream>
#include<cmath>
#include<algorithm>
#include<queue>
#define ll long long
#define pii pair<int,int>
#define PINF 0x7fffffff
#define NINF 0x80000000
using namespace std;
int s, p;
struct edge {
int u, v;
double w;
bool operator < (const edge& e2) const {
return w < e2.w;
}
}edges[125000];
struct point {
int x, y;
}points[505];
int father[505];
int _find(int x) {
if (father[x] == 0)return x;
return father[x] = _find(father[x]);
}
bool _union(int x, int y) {
x = _find(x); y = _find(y);
if (x == y)return false;
father[x] = y;
return true;
}
inline double dis(int x1, int y1, int x2, int y2) {
return sqrt((double)((x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2)));
}
int main() {
//ios::sync_with_stdio(false);
cin >> s >> p;
for (int i = 1; i <= p; i++) {
cin >> points[i].x >> points[i].y;
}
int ePtr = 0;
for (int i = 1; i <= p; i++) {
for (int j = i + 1; j <= p; j++) {
ePtr++;
edges[ePtr].u = i;
edges[ePtr].v = j;
edges[ePtr].w = dis(points[i].x, points[i].y, points[j].x, points[j].y);
}
}
sort(edges + 1, edges + ePtr + 1);
int cnter = 0;
for (int i = 1; i <= ePtr; i++) {
if (_union(edges[i].u, edges[i].v)) {
cnter++;
}
if (cnter == p - 1 - (s - 1)) {
printf("%.2lf", edges[i].w);
break;
}
}
}

杰克部落

梦想本不会发光，发光的是追梦的你

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Openjudge BJUTACM 19g系列代码

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洛谷 P3376 【模板】网络最大流

洛谷 P3386 【模板】二分图匹配

计蒜客 Random Access Iterator

洛谷 P1113 杂务

洛谷 P1144 最短路计数

洛谷 P1346 电车

洛谷 P1991 无线通讯网

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