HDU_3313_Key_Vertex(最小割)

题意：已给有向图，给定起点和终点S，T，问图中有多少个点去掉可以使S T不可达

思路：

按常规思路就是找一条最短路，最短路之外的所有点不符合条件，然后在最短路上搞搞就好了

为了加深一下最小割的理解就用最小割写了此题

拆点，拆点之间连容量为1的边，有向边连inf的边，最大流一下，显然关键点的拆点间的流是满流而且必须是最小割。

然后就从满流+最小割这两点判断就可以辣，若u -> v 不是最小割，那么残余网络上S必然可达v，所以在残余网络上bfsS的可达点。

/* **********************************************
Auther: kalili
Created Time: 2015/6/17 18:10:30
File Name   : HDU_3313_Key_Vertex最小割.cpp
*********************************************** */
//Accepted 3313 2386MS 19136K 5692 B C++
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <sstream>
#include <string>
#include <vector>
#include <cstdio>
#include <ctime>
#include <bitset>
#include <algorithm>
#define SZ(x) ((int)(x).size())
#define ALL(v) (v).begin(), (v).end()
#define foreach(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); ++ i)
#define REP(i,n) for ( int i=1; i<=int(n); i++ )
using namespace std;

const int inf = 0x3f3f3f3f;
const int N = 200000 +100;
const int M = 300000*10+10;
struct Edge
{
        int v,w,nxt;
        Edge(){}
        Edge(int v,int w,int nxt) : v(v),w(w),nxt(nxt) {}
}es[M];
int n,m;
int head[N];
int cnt;
void inline add_edge(int u,int v,int w)
{
        es[cnt]=Edge(v,w,head[u]);
        head[u]=cnt++;
        es[cnt]=Edge(u,0,head[v]);
        head[v]=cnt++;
}
int s,g;

int h[N],q[N],tail;
bool makeh(int s,int g)
{
        memset(h,0,sizeof(h));
        h[s]=1;
        tail=0;
        q[tail++]=s;
        for(int i=0;i<tail;i++)
        {
                int u = q[i];
                if(u ==  g) return true;
                for(int i = head[u];~i;i = es[i].nxt)
                {
                        int v = es[i].v,w = es[i].w;
                        if(w&&h[v] == 0) q[tail++]=v,h[v] = h[u]+1;
                }
        }
        return false;
}
int dfs(int u,int g,int maxf)
{
        if(u == g) return maxf;
        int ans=0;
        for(int i = head[u];~i;i = es[i].nxt)
        {
                int v = es[i].v,w = es[i].w;
                if(w&&h[v] == h[u]+1)
                {
                        int f = dfs(v,g,min(maxf-ans,w));
                        ans += f;
                        es[i].w -= f;
                        es[i^1].w += f;
                        if(ans == maxf) return ans;
                }
        }
        if(ans == 0) h[u]=-1;
        return ans;
}
int Dinic(int s,int g)
{
        int ans=0;
        while(makeh(s,g)) ans+=dfs(s,g,inf);
        return ans;
}
bool vis[N];
int que[N],top;
void bfs(int src)
{
        que[top++] = src;
        vis[src]=1;
        for(int i = 0;i < top; i++)
        {
                int u=que[i];
                for(int i = head[u]; ~i ;i = es[i].nxt)
                {
                        int v = es[i].v,w = es[i].w;
                        if(w&&vis[v] == 0) que[top++]=v,vis[v]=1;
                }
        }
}
void ini()
{
        cnt=0;
        memset(head,-1,sizeof(head));
}
int main()
{
        while(~scanf("%d%d",&n,&m))
        {
                ini();
                REP(i,m)
                {
                        int u,v;
                        scanf("%d%d",&u,&v);
                        add_edge(u+n,v,inf);
                }
                scanf("%d%d",&s,&g);
                for(int i=0;i<n;i++)
                {
                        if(i==s||i==g) add_edge(i,i+n,2);
                        else add_edge(i,i+n,1);
                }
                int ans=Dinic(s,g+n);
                if(ans == 2) puts("2");
                if(ans == 0) printf("%d\n",n);
                if(ans == 1)
                {
                        int cnt=0;
                        int src=s;
                        es[head[s]].w = es[head[g]].w=0;
                        memset(vis,0,sizeof(vis));
                        while(true)
                        {
                                top=0;
                                bfs(src);
                                bool flag=0;
                                for(int i =0 ; i < top; i++)
                                {
                                        int u=que[i];
                                        if(flag) break;
                                        for(int i = head[u];~i;i = es[i].nxt)
                                        {
                                                if(i&1) continue;
                                                int v=es[i].v,w=es[i].w;
                                                if(w == 0 && vis[v] == 0)
                                                {
                                                        cnt++;
                                                        src = v;
                                                        flag=1;
                                                        break;
                                                }
                                        }
                                }
                                if(src==g+n) break;
                        }
                        printf("%d\n",cnt);
                }

        }

}