台州学院第五届大学生程序设计竞赛

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扔硬币

Solve the Equation

视频监控

MFA Algorithm

Circle VS Triangle

电脑密码

The Key Locker of Cell Phone

Number Maze

修路问题

浪漫自习

最大流EK 可以过

取一个源点0 建一条边 0-1 容量为m(情侣的对数)

如果最大流等于m 就YES

此外我建图比较麻烦

对于每一条无向边我拆成2条有向每条有向边还有一条反向边 4条初学

#include <cstdio>#include <cstring>#include <queue>using namespace std;const int MAX = 55;int n,m;struct edge{int u;int v;int flow;int cap;int next;}e[10000];int cap[MAX][MAX];int a[MAX];int first[MAX];int p[MAX];int edgenum;int f,c;int s,t;void add(int u,int v,int cap){e[edgenum].u = u;e[edgenum].v = v;e[edgenum].flow = 0;e[edgenum].cap = cap;e[edgenum].next = first[u];first[u] = edgenum++;e[edgenum].u = v;e[edgenum].v = u;e[edgenum].flow = 0;e[edgenum].cap = 0;e[edgenum].next = first[v];first[v] = edgenum++;}void EK(){queue <int> q;f = 0;while(1){memset(a,0,sizeof(a));memset(p,-1,sizeof(p));a[0] = 999999999;q.push(0);while(!q.empty()){int u = q.front();q.pop();for(int k = first[u]; k != -1; k = e[k].next){int v = e[k].v;if(!a[v] && e[k].cap > e[k].flow){p[v] = k;q.push(v);a[v] = min(a[u],e[k].cap - e[k].flow);}}}if(a[t] == 0)break;for(int k = p[t]; k != -1; k = p[e[k].u]){e[k].flow += a[t];e[k^1].flow -= a[t];}f += a[t];}}int main(){int i,j,k;while(scanf("%d %d",&m,&n),n||m){edgenum = 0;s = 0;t = 2;memset(first,-1,sizeof(first));memset(cap,0,sizeof(cap));add(0,1,m);for(i = 1; i <= n; i++){scanf("%d",&k);while(k--){scanf("%d",&j);cap[i][j] = cap[j][i] = 1;//add(i,j,1);//add(j,i,1);}}for(i = 1;i <= n; i++)for(j = i+1; j <= n; j++)if(cap[i][j]){add(i,j,1);add(j,i,1);}EK();if(f == m)printf("YES\n");elseprintf("NO\n");}return 0;}

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