洛谷 1344 [USACO4.4]追查坏牛奶Pollutant Control——最大流

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

那个边数的限制，只要把边权乘1001再+1即可。乘1001是因为有1000条边，这样流量小的不会因为边数多而被认为不优。不是乘1000是为了/1001和%1001取出答案,1000的话略有冲突。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
using namespace std;
const int N=35,M=1005;
const ll INF=0x3f3f3f3f3f3f3f3f;
int n,m,hd[N],cur[N],xnt=1,dfn[N];
int q[N],he,tl;
ll mxflow;
struct Ed{
    int nxt,to,cap;
    Ed(int n=0,int t=0,int c=0):nxt(n),to(t),cap(c) {}
}ed[M<<1];
void add(int x,int y,int z)
{
    ed[++xnt]=Ed(hd[x],y,z);hd[x]=xnt;
    ed[++xnt]=Ed(hd[y],x,0);hd[y]=xnt;
}
bool bfs()
{
    memset(dfn,0,sizeof dfn);
    dfn[1]=1; he=tl=0; q[++tl]=1;
    while(he<tl)
    {
        int k=q[++he];
        for(int i=hd[k],v;i;i=ed[i].nxt)
            if(ed[i].cap&&!dfn[v=ed[i].to])
                dfn[v]=dfn[k]+1,q[++tl]=v;
    }
    return dfn[n];
}
ll dinic(int cr,ll flow)
{
    if(cr==n) return flow;
    ll use=0;
    for(int &i=cur[cr],v;i;i=ed[i].nxt)
        if(dfn[v=ed[i].to]==dfn[cr]+1&&ed[i].cap)
        {
            ll tmp=dinic(v,min(flow-use,(ll)ed[i].cap));
            if(!tmp) dfn[v]=0;
            ed[i].cap-=tmp; ed[i^1].cap+=tmp;
            use+=tmp; if(use==flow) return use;
        }
    return use;
}
int main()
{
    scanf("%d%d",&n,&m);
    for(int i=1,u,v,z;i<=m;i++)
    {
        scanf("%d%d%d",&u,&v,&z);
        add(u,v,z*1001+1);
    }
    while(bfs())
    {
        memcpy(cur,hd,sizeof hd);
        mxflow+=dinic(1,INF);
    }
    printf("%lld %lld\n",mxflow/1001,mxflow%1001);
    return 0;
}