网络最大流最短增广路Dinic算法模板

//1是源点，n是汇点。

#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

const int maxn=1000+10;
const int INF=0x7FFFFFFF;

struct Edge { int from,to,cap,flow; };
vector<Edge>edges;
vector<int>G[maxn];
bool vis[maxn];
int d[maxn];
int cur[maxn];
int n,m,s,t;

bool BFS()
{
    memset(vis,0,sizeof(vis));
    queue<int>Q;
    Q.push(s);
    d[s]=0;
    vis[s]=1;
    while(!Q.empty())
    {
        int x=Q.front();
        Q.pop();
        for(int i=0;i<G[x].size();i++)
        {
            Edge& e=edges[G[x][i]];
            if(!vis[e.to]&&e.cap>e.flow)
            {
                vis[e.to]=1;
                d[e.to]=d[x]+1;
                Q.push(e.to);
            }
        }
    }
    return vis[t];
}

void AddEdge(int from,int to,int cap)
{
    Edge r;
    r.from=from;r.to=to;r.cap=cap;r.flow=0;
    edges.push_back(r);
    Edge d;
    d.from=to;d.to=from;d.cap=0;d.flow=0;
    edges.push_back(d);
    m=edges.size();
    G[from].push_back(m-2);
    G[to].push_back(m-1);
}

int DFS(int x,int a)
{
    if(x==t||a==0) return a;
    int flow=0,f;
    for(int& i=cur[x];i<G[x].size();i++)
    {
        Edge& e=edges[G[x][i]];
        if(d[x]+1==d[e.to]&&(f=DFS(e.to,min(a,e.cap-e.flow)))>0)
        {
            e.flow+=f;
            edges[G[x][i]^1].flow-=f;
            flow+=f;
            a-=f;
            if(a==0) break;
        }
    }
    return flow;
}

int Maxflow(int ss,int tt)
{
    int flow=0;
    while(BFS())
    {
        memset(cur,0,sizeof(cur));
        flow+=DFS(ss,INF);
    }
    return flow;
}

int main()
{
    int T,ii;
    scanf("%d",&T);
    for(ii=1;ii<=T;ii++)
    {
        edges.clear();
        for(int i=0;i<maxn;i++) G[i].clear();
        int mm;
        scanf("%d%d",&n,&mm);
        s=1;t=n;
        while(mm--)
        {
            int uu,vv,cc;
            scanf("%d%d%d",&uu,&vv,&cc);
            AddEdge(uu,vv,cc);
        }
        printf("Case %d: %d\n",ii,Maxflow(s,t));
    }
    return 0;
}