BZOJ 2502: 清理雪道 [最小流]

2502: 清理雪道

题意：任意点出发任意次每条边至少经过一次最小花费。

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下界1，裸最小流....

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define fir first
#define sec second
typedef long long ll;
const int N=1005, M=4e5+5, INF=1e9;
inline ll read(){
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, m, x, s, t, tot, extra[N];
struct edge{int v, c, f, ne, lower;}e[M];
int cnt=1, h[N];
inline int ins(int u, int v, int c, int b=0) {
	e[++cnt]=(edge){v, c, 0, h[u], b}; h[u]=cnt;
	e[++cnt]=(edge){u, 0, 0, h[v], b}; h[v]=cnt;
	return cnt-1;
}
int q[N], head, tail, vis[N], d[N], cur[N];
bool bfs(int s, int t) {
	memset(vis, 0, sizeof(vis));
	head=tail=1;
	q[tail++]=s; d[s]=0; vis[s]=1;
	while(head!=tail) {
		int u=q[head++];
		for(int i=h[u];i;i=e[i].ne) 
			if(!vis[e[i].v] && e[i].c>e[i].f) { 
				vis[e[i].v]=1; d[e[i].v]=d[u]+1;
				q[tail++]=e[i].v;
				if(e[i].v == t) return true;
			}
	}
	return false;
}
int dfs(int u, int a, int t) { 
	if(u==t || a==0) return a;
	int flow=0, f;
	for(int &i=cur[u];i;i=e[i].ne) 
		if(d[e[i].v]==d[u]+1 && (f=dfs(e[i].v, min(a, e[i].c-e[i].f), t))>0) {
			flow+=f;
			e[i].f+=f;
			e[i^1].f-=f;
			a-=f;
			if(a==0) break;
		}
	if(a) d[u]=-1;
	return flow;
}
int dinic(int s, int t) {
	int flow=0;
	while(bfs(s, t)) {
		for(int i=0; i<=tot; i++) cur[i]=h[i];
		flow+=dfs(s, INF, t); 
	}
	return flow;
}

int main() {
	freopen("in","r",stdin);
	n=read(); s=0; t=n+1;
	for(int i=1; i<=n; i++) {
		m=read();
		while(m--) x=read(), ins(i, x, INF, 1), extra[i]--, extra[x]++;
		ins(s, i, INF); ins(i, t, INF);
	}
	int ss=t+1, tt=t+2, sum=0; tot=tt;
	for(int i=s; i<=t; i++) {
		if(extra[i]>0) ins(ss, i, extra[i]), sum+=extra[i];
		if(extra[i]<0) ins(i, tt, -extra[i]);
	}
	ins(t, s, INF);
	int flow=dinic(ss, tt); //printf("flow %d\n",flow);
	int feas = e[cnt-1].f;
	e[cnt-1].c = e[cnt].c = 0;
	printf("%d",feas-dinic(t, s));
}