No.36 - POJ1966 图的联通度-枚举T网络流

求一个图的联通度，就是找图中固定源点，然后枚举汇点求最小割。
然后最小割中最小的一个就是图的联通度。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<vector>
#include<math.h>
#define MM(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;

/*
    ShellDawn
    POJ1966
    No.36
 */

#define maxn 55
int N,M;
int S,T;
struct Edge{
    int to;
    int v;
    int rvs;
    int pre;
};
Edge E[maxn*maxn*4];
int pre[maxn*2];
int cnt;

int In[maxn*maxn][2];
int V[maxn*2];

void addE(int a,int b,int v){
    E[cnt].to = b;
    E[cnt].v = v;
    E[cnt].rvs = cnt+1;
    E[cnt].pre = pre[a];
    pre[a] = cnt++;
    E[cnt].to = a;
    E[cnt].v = 0;
    E[cnt].rvs = cnt - 1;
    E[cnt].pre = pre[b];
    pre[b] =  cnt ++;
}

bool BFS(int s){
    MM(V,0);
    queue<int> q;
    q.push(s);
    V[s] = 1;
    while(!q.empty()){
        int now = q.front();
        q.pop();
        for(int i=pre[now];i!=0;i=E[i].pre){
            if(E[i].v > 0 && V[E[i].to] == 0){
                V[E[i].to] = V[now] + 1;
                q.push(E[i].to);
            }
        }
    }
    if(V[T] > 0) return true;
    return false;
}

int DFS(int now,int minflow){
    if(now == T) return minflow;
    int flow = 0;
    for(int i=pre[now];i!=0 && minflow; i = E[i].pre){
        if(E[i].v > 0 && V[E[i].to] == V[now] + 1){
            int f = DFS(E[i].to,min(minflow,E[i].v));
            flow += f;
            minflow -= f;
            E[i].v -= f;
            E[E[i].rvs].v += f;
        }
    }
    if(flow == 0) V[now] = 0;
    return flow;
}

void build(){
    MM(pre,0); cnt = 1;
    for(int i=0;i<N;i++) addE(i,i+N,1);
    for(int i=0;i<M;i++){
        int a,b;
        a = In[i][0];
        b = In[i][1];
        addE(a+N,b,INF);
        addE(b+N,a,INF);
    }
}

int main(){
    //freopen("A.txt","r",stdin);
    while(~scanf("%d%d",&N,&M)){
        if(N == 0) { printf("0\n");continue; }
        if(M == 0 && N == 1) { printf("1\n");continue;}
        if(M == 0 && N > 1){printf("0\n");continue;}
        for(int i=0;i<M;i++) scanf(" (%d,%d)",&In[i][0],&In[i][1]);
        int ans = INF;
        S = N;
        for(int i=1;i<N;i++){
            T = i;
            build();
            int maxflow = 0;
            while(BFS(S)) maxflow += DFS(S,INF);
            ans = min(ans,maxflow);
        }
        printf("%d\n",ans>=INF?N:ans);
    }
    return 0;
}