POJ 1966 Cable TV Network | 最小點割集、最小割

這題簡直亂搞。。但凡最小點割集，均需拆點。

題意：

給你一個圖，讓你求出圖中的最小點割集裡包含點的個數。

思路：

暴力枚舉。任選兩個點作為s，t。然後跑最大流。

取所有最大流中最小的那個值就是本題答案。

建圖：

拆點：每個點拆成兩個點，容量為1。

AC代碼：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <queue>
#include <vector>
#define debug cout<<"debug"<<endl;
using namespace std;
const int MAXEDGE = 5*1e5 + 5;
const int INF = 0x3f3f3f3f;
const int MAXN = 150;
int n, m;
struct Edge
{
        int to, cap, flow, next;
}edge[MAXEDGE], backUp[MAXEDGE];
int backUpHead[MAXN];
int savePP;
struct Dinic
{
        int s, t, pp;
        int head[MAXN], cur[MAXN], d[MAXN];
        bool vis[MAXN];
        Dinic()
        {
                pp = 0;
                memset(head, -1, sizeof(head));
        }
        void addEdge(int u, int v, int c)
        {
                edge[pp] = (Edge){v, c, 0, head[u]};
                backUp[pp] = (Edge){v, c, 0, head[u]};
                head[u] = pp++;
                backUpHead[u] = head[u];

                edge[pp] = (Edge){u, 0, 0, head[v]};
                backUp[pp] = (Edge){u, 0, 0, head[v]};
                head[v] = pp++;
                backUpHead[v] = head[v];
        }
        void addEdge2(int u, int v, int c)
        {
                edge[pp] = (Edge){v, c, 0, head[u]};
                head[u] = pp++;

                edge[pp] = (Edge){u, 0, 0, head[v]};
                head[v] = pp++;

        }
        bool bfs()
        {
                memset(vis, false, sizeof(vis));
                d[s] = 0;
                queue <int> q;
                q.push(s);
                vis[s] = true;
                while(!q.empty())
                {
                        int u = q.front();
                        q.pop();
                        int next = head[u];
                        while(next != -1)
                        {
                                Edge &e = edge[next];
                                if(!vis[e.to] && e.cap > e.flow)
                                {
                                        q.push(e.to);
                                        vis[e.to] = true;
                                        d[e.to] = d[u] + 1;
                                        //cout<<"e.to = "<<e.to<<endl;
                                }
                                next = e.next;
                        }
                }
                return vis[t];
        }
        int dfs(int u, int a)
        {
                if(u == t || a == 0)    return a;
                int flow = 0, f;
                int &next = cur[u];
                while(next != -1)
                {
                        Edge &e = edge[next];
                        if(d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0)
                        {
                                e.flow += f;
                                edge[next^1].flow -= f;
                                flow += f;
                                a -= f;
                                if(a == 0)      break;
                        }
                        next = e.next;
                }
                return flow;
        }
        void recover()
        {
                for(int i = 0;i < pp; i++)
                {
                        edge[i] = backUp[i];
                }
                for(int i = 0;i < n*2 + 5; i++)
                {
                        head[i] = backUpHead[i];
                }
        }
        int getMaxFlow()
        {
                int res = 0;
                while(bfs())
                {
                        for(int i = 0;i < 2*n + 2; i++)
                        {
                                cur[i] = head[i];
                        }
                        res += dfs(s, INF);
                }
                return res;
        }
};

int a[MAXN][MAXN];
int main()
{        while(scanf("%d%d",&n ,&m) != EOF)
        {
                int u, v;
                Dinic dinic;
                for(int i = 0; i < n; i++)
                {
                        dinic.addEdge(i+n, i, 1);
                }
                for(int i = 0;i < m; i++)
                {
                        scanf(" (%d,%d)",&u ,&v);
                        dinic.addEdge(u, v + n, INF);
                        dinic.addEdge(v, u + n, INF);
                }
                savePP = dinic.pp;
                //special

                int s, t;
                int res = n;
                for(int i = 0;i < n; i++)
                {
                        for(int j = 0;j < n; j++)
                        {
                                if(i == j)      continue;
                                dinic.s = i+n, dinic.t = j;
                                dinic.addEdge2(i + n, i, INF);
                                dinic.addEdge2(j + n, j, INF);
                                //cout<<"1"<<endl;
                                res = min(res, dinic.getMaxFlow());
                                dinic.pp = savePP;
                                dinic.recover();
                        }
                }
                printf("%d\n",res);
        }
        return 0;
}