【USACO4.4】追查坏牛奶

题面

https://www.luogu.org/problem/P1344

题解

$gyfan$的神仙题。

给每个边一个无限小值的就行了。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#define ri register int
#define N 20500
#define INF 1000000007
using namespace std;
vector<int> ed[N];
int w[N<<1],to[N<<1];
int n,m,cnt;
long long ans;
int vis[N],d[N],cur[N];
queue<int> q;

void add_edge(int a,int b,int c) {
  ++cnt; w[cnt]=c; to[cnt]=b; ed[a].push_back(cnt);
  ++cnt; w[cnt]=0; to[cnt]=a; ed[b].push_back(cnt);
}

bool bfs(){
  vis[1]=1;
  q.push(1);
  while (!q.empty()) {
    int x=q.front(); q.pop();
    for (ri i=0;i<ed[x].size();i++) {
      int e=ed[x][i];
      if (vis[to[e]] || !w[e]) continue;
      vis[to[e]]=1;
      d[to[e]]=d[x]+1;
      q.push(to[e]);
    }
  }
  return vis[n];
}

int dfs(int x,int lim){
  int tl=lim;
  if (x==n || !lim) return lim;
  for (ri i=cur[x];i<ed[x].size();i++) {
    int e=ed[x][i];
    if (!w[e] || d[to[e]]!=d[x]+1) continue;
    int f=dfs(to[e],min(lim,w[e]));
    if (!f) continue;
    w[e]-=f; w[1^e]+=f; lim-=f;
    if (!lim) return tl;
    cur[x]=i;
  }
  return tl-lim;
}

void dinic(){
  while (bfs()) {
    for (ri i=0;i<=n;i++) vis[i]=0;
    for (ri i=0;i<=n;i++) cur[i]=0;
    while (1) {
      int t=dfs(1,INF);
      if (!t) break; else ans+=t;
    }
  }
}

int main(){
  scanf("%d %d",&n,&m);
  cnt=-1;
  for (ri i=1;i<=m;i++) {
    int a,b,c;
    scanf("%d %d %d",&a,&b,&c);
    add_edge(a,b,c*1010+1);
  }
  ans=0LL;
  dinic();
  cout<<ans/1010<<" "<<ans%1010<<endl;
}