网络流24题3

最小路径覆盖

https://loj.ac/problem/6002

因为每个点只能用一次，因此前驱和后继只能连一个点，拆点，对于一条u，v的边连u作为前驱，v作为后继的点

然后源点连上所有前驱点

汇点连上所有后继点

跑最大流解二分图匹配问题

因为每有有个匹配 说明点u和v从2个路径接成一个路径了，使得总路径数-1，因此答案等于点数-最大流量

#include <iostream>
#include <cstring>
#include <stdio.h>
#include <queue>
using namespace std;
const int maxn = 6009;
int tot = 0;
int head[maxn];
struct edge{
    int v,nex,w;
}e[maxn*2];
void addedge(int u,int v,int w){
    e[tot] = (edge){v,head[u],w};
    head[u] = tot++;
    e[tot] = (edge){u,head[v],0};
    head[v] = tot++;
}
int deep[maxn];
bool bfs(int S,int T){
    memset(deep,0,sizeof(deep));
    deep[S] = 1;
    queue<int> q;
    q.push(S);
    while(!q.empty()){
        int now = q.front();
        q.pop();
        for(int i=head[now];i!=-1;i=e[i].nex){
            int v = e[i].v;
            int w = e[i].w;
            if(deep[v]!=0 || w<=0) continue;
            deep[v] = deep[now]+1;
            q.push(v);
        }
    }
    return deep[T];
}
int dfs(int now,int T,int maxflow){
    if(now==T) return maxflow;
    int all = 0;
    for(int i=head[now];i!=-1 && all<maxflow;i=e[i].nex){
        int v = e[i].v;
        int w = e[i].w;
        if(deep[v]!=deep[now]+1 || w<=0) continue;
        int tt = dfs(v,T,min(maxflow-all,w));
        e[i].w-=tt;
        e[i^1].w+=tt;
        all+=tt;
    }
    return all;
}
int dinic(int S,int T){
    int ret = 0;
    while(bfs(S,T)){
        ret+=dfs(S,T,0x3f3f3f3f);
    }
    return ret;
}
int vis[maxn];
void print(int now){
    if(now<=1) return;
    vis[now] = 1;
    printf("%d ",now/2);
    for(int i=head[now];i!=-1;i=e[i].nex){
        int v = e[i].v;
        int w = e[i].w;
        if(w<=0){
            print(v^1);
        }
    }
}
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    memset(head,-1,sizeof(head));
    for(int i=1;i<=m;i++){
        int u,v;
        scanf("%d%d",&u,&v);
        addedge(2*u,2*v+1,1);
    }
    for(int i=1;i<=n;i++){
        addedge(0,2*i,1);
        addedge(2*i+1,1,1);
    }
    int ans = dinic(0,1);
    for(int i=1;i<=n;i++){
        if(vis[2*i]==0) {
            print(2*i);
            printf("\n");
        }
    }
    printf("%d\n",n-ans);
    return 0;
}