洛谷 P1344 [USACO4.4]追查坏牛奶Pollutant Control

https://www.luogu.org/problemnew/show/P1344

题目要求最小割和边数最小的最小割的边数

前者容易，重点是后者

方法：https://blog.csdn.net/chenzhenyu123456/article/details/48103229

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<iostream>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
namespace F
{

struct E
{
    int to,nxt,from,cap,flow;
}e[400100];
int f1[60100],ne=1;
int S,T,n;
int d[60100];
bool bfs()
{
    int k,u;
    memset(d,0,sizeof(int)*(n+1));
    queue<int> q;
    q.push(S);d[S]=1;
    while(!q.empty())
    {
        u=q.front();q.pop();
        for(k=f1[u];k;k=e[k].nxt)
            if(!d[e[k].to]&&e[k].cap>e[k].flow)
            {
                d[e[k].to]=d[u]+1;
                //if(e[k].to==T)    return 1;
                q.push(e[k].to);
            }
    }
    //return 0;
    return d[T];
}
int cur[60100];
int dfs(int u,int x)
{
    if(u==T||x==0)    return x;
    int flow=0,f;
    for(int &k=cur[u];k;k=e[k].nxt)
        if(e[k].cap>e[k].flow&&d[e[k].to]==d[u]+1)
        {
            f=dfs(e[k].to,min(x-flow,e[k].cap-e[k].flow));
            e[k].flow+=f;e[k^1].flow-=f;flow+=f;
            if(flow==x)    return flow;
        }
    return flow;
}
int solve()
{
    int flow=0;
    while(bfs())
    {
        memcpy(cur,f1,sizeof(int)*(n+1));
        flow+=dfs(S,0x3f3f3f3f);
    }
    return flow;
}
void me(int a,int b,int c)
{
    e[++ne].to=b;e[ne].nxt=f1[a];f1[a]=ne;
    e[ne].from=a;e[ne].cap=c;e[ne].flow=0;
    e[++ne].to=a;e[ne].nxt=f1[b];f1[b]=ne;
    e[ne].from=b;e[ne].cap=0;e[ne].flow=0;
}


}

int n,m;
int main()
{
    int i,a,b,c;
    scanf("%d%d",&n,&m);F::n=n;F::S=1;F::T=n;
    for(i=1;i<=m;i++)
    {
        scanf("%d%d%d",&a,&b,&c);
        F::me(a,b,c);
    }
    int an=F::solve();
    for(i=2;i<=F::ne;i+=2)
    {
        using F::e;
        if(e[i].cap==e[i].flow)
        {
            e[i].cap=1;e[i^1].cap=0;
        }
        else
        {
            e[i].cap=0x3f3f3f3f;e[i^1].cap=0;
        }
        e[i].flow=0;e[i^1].flow=0;
    }
    printf("%d %d",an,F::solve());
    return 0;
}