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));
}