Smallest Minimum Cut HDU - 6214 最少边的最小割

结论题，加边的时候每个边容量扩大为cap*(Edge+1)+1，跑一边最大流%(Edge+1)就是最少边数了

证明一下就是一个容量为8的边和两个容量为4的边，扩大后只有堵住原先8才是最小的

新最大流：maxflow*(Edge+1)+最少割边数

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=2e5+7;
const int inf=0x3f3f3f3f;

int num_nodes;
int tot;
int d[maxn],cur[maxn],source,sink;
int p[maxn],num[maxn],vis[maxn];

struct EDGE{
    int from,to,cap,flow;
};
vector<EDGE>edge;
vector<int>G[maxn];

void init(){
    edge.clear();
    for(int i=0;i<maxn;i++)G[i].clear();
}

void addedge(int u,int v,int w){
    edge.push_back((EDGE){u,v,w,0});
    edge.push_back((EDGE){v,u,0,0});
    tot=edge.size();
    G[u].push_back(tot-2);
    G[v].push_back(tot-1);
}

int bfs(){
    memset(vis,0,sizeof vis);
    queue<int>Q;
    Q.push(sink);
    vis[sink]=1;
    d[sink]=0;
    while(!Q.empty()){
        int u=Q.front();
        Q.pop();
        int sz=G[u].size();
        for(int i=0;i<sz;++i){
            EDGE &e=edge[G[u][i]^1];
            if(!vis[e.from] && e.cap>e.flow){
                vis[e.from]=1;
                d[e.from]=d[u]+1;
                Q.push(e.from);
            }
        }
    }
    return vis[source];
}

int augment(){
    int u=sink,a=inf;
    while(u!=source){
        EDGE &e=edge[p[u]];
        a=min(a,e.cap-e.flow);
        u=edge[p[u]].from;
    }
    u=sink;
    while(u!=source){
        edge[p[u]].flow+=a;
        edge[p[u]^1].flow-=a;
        u=edge[p[u]].from;
    }
    return a;
}

int maxflow(){
    int flow=0;
    bfs();
    memset(num,0,sizeof num);
    for(int i=0;i<num_nodes;i++)num[d[i]]++;
    int u=source;
    memset(cur,0,sizeof cur);
    while(d[source]<num_nodes){
        if(u==sink){
            flow+=augment();
            u=source;
        }
        int advance=0;
        int sz=G[u].size();
        for(int i=cur[u];i<sz;i++){
            EDGE &e=edge[G[u][i]];
            if(e.cap>e.flow && d[u]==d[e.to]+1){
                advance=1;
                p[e.to]=G[u][i];
                cur[u]=i;
                u=e.to;
                break;
            }
        }
        if(!advance){
            int m=num_nodes-1;
            int sz=G[u].size();
            for(int i=0;i<sz;i++){
                if(edge[G[u][i]].cap>edge[G[u][i]].flow){
                    m=min(m,d[edge[G[u][i]].to]);
                }            
            }
            if(--num[d[u]]==0)break;
            num[d[u]=m+1]++;
            cur[u]=0;
            if(u!=source)u=edge[p[u]].from;
        }
    }
    return flow;
}

int main(){
    int T;scanf("%d",&T);
    while(T--){
        init();
        int n,m;scanf("%d%d",&n,&m);
        num_nodes=n+10;
        scanf("%d%d",&source,&sink);
        for(int i=1,u,v,w;i<=m;i++){
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w*(m+1)+1);//addedge(v,u,w*(m+1)+1);
        }
        printf("%d\n",maxflow()%(m+1));
    }
}