最大流EK演算法模板

#include<cstdio>        //EK()演算法。時間複雜度（VE^2）#include<queue>#include<cstring>using namespace std;const int maxn = 100;const int INF = (1<<30)-1;int g[maxn][maxn];int flow[maxn],pre[maxn];bool vis[maxn];int n,m;int bfs(int s,int e){    memset(pre,-1,sizeof(pre));    memset(vis,false,sizeof(vis));    queue<int > q;    vis[s] = true;    for(int i=1;i<=n;i++)   flow[i]=INF;    q.push(s);    while(!q.empty()){        int now = q.front();  q.pop();        if(now==n)  break;        for(int i=1;i<=n;i++){                //尋找增廣路最小流量            if(!vis[i]&&g[now][i]>0){                vis[i] = true;                flow[i] = min(flow[now],g[now][i]);                pre[i] = now;                q.push(i);            }        }    }    if(!vis[e]|| e==1)                         //找不到完整的增廣路or源點匯點重合         return -1;    else         return flow[e];}int EK(int s,int e){    int temp,d,res,maxflow;    maxflow = 0;    while((d=bfs(s,e))!=-1){         maxflow += d;         temp=n;         while(temp!=1){             res = pre[temp];             g[res][temp]-=d;               //正向邊             g[temp][res]+=d;               //反向邊             temp = res;         }    }    return maxflow;}int main(){    int T,ca=1;    int start,end,capacity;    scanf("%d",&T);    while(T--){        memset(g,0,sizeof(g));        scanf("%d %d",&n,&m);        for(int i=1;i<=m;i++){            scanf("%d%d%d",&start,&end,&capacity);            g[start][end]+=capacity;        }        printf("Case %d: %d\n",ca++,EK(1,n));    }    return 0;}