• [POJ 1966] Cable TV Network


    [题目链接]

             http://poj.org/problem?id=1966

    [算法]

             拆点 + 最小割

    [代码]

             

    #include <algorithm>  
    #include <bitset>  
    #include <cctype>  
    #include <cerrno>  
    #include <clocale>  
    #include <cmath>  
    #include <complex>  
    #include <cstdio>  
    #include <cstdlib>  
    #include <cstring>  
    #include <ctime>  
    #include <deque>  
    #include <exception>  
    #include <fstream>  
    #include <functional>  
    #include <limits>  
    #include <list>  
    #include <map>  
    #include <iomanip>  
    #include <ios>  
    #include <iosfwd>  
    #include <iostream>  
    #include <istream>  
    #include <ostream>  
    #include <queue>  
    #include <set>  
    #include <sstream>  
    #include <stdexcept>  
    #include <streambuf>  
    #include <string>  
    #include <utility>  
    #include <vector>  
    #include <cwchar>  
    #include <cwctype>  
    #include <stack>  
    #include <limits.h>
    using namespace std;
    const int inf = 2e9;
    #define MAXN 55
    
    int i,n,m,s,t,ans,S,T,tot;
    int head[MAXN << 1],depth[MAXN << 1],u[MAXN * MAXN],v[MAXN * MAXN];
    bool g[MAXN][MAXN];
    
    struct edge
    {
            int to,w,nxt;
    } e[MAXN * MAXN * 2];
    
    inline void addedge(int u,int v,int w)
    {
            tot++;
            e[tot] = (edge){v,w,head[u]};
            head[u] = tot;
            tot++;
            e[tot] = (edge){u,0,head[v]};
            head[v] = tot;
    }
    inline bool bfs()
    {
            int i,l,r,u,v,w;
            static int q[MAXN << 1];
            memset(depth,0,sizeof(depth));
            q[l = r = 1] = S;
            depth[S] = 1;
            while (l <= r)
            {
                    u = q[l];
                    l++;
                    for (i = head[u]; i; i = e[i].nxt)
                    {
                            v = e[i].to;
                            w = e[i].w;
                            if (!depth[v] && w)
                            {
                                    depth[v] = depth[u] + 1;
                                    q[++r] = v;
                                    if (v == T) return true;
                            }
                    }
            }
            return false;
    }
    inline int dinic(int u,int flow)
    {
            int i,v,w,k,rest = flow;
            if (u == T) return flow;
            for (i = head[u]; i && rest; i = e[i].nxt)
            {
                    v = e[i].to;
                    w = e[i].w;
                    if (depth[v] == depth[u] + 1 && w)
                    {
                            k = dinic(v,min(rest,w));
                            if (!k) depth[v] = 0;
                            e[i].w -= k;
                            e[i ^ 1].w += k;
                            rest -= k;
                    }
            }
            return flow - rest;
    }
    inline int getans(int s,int t)
    {
            int i,Minimum_Cut,flow;
            tot = 1;
            memset(head,0,sizeof(head));
            for (i = 0; i < n; i++) 
            {
                    if (i != s && i != t)
                            addedge(i,i + n,1);
                    else addedge(i,i + n,inf);
            }
            for (i = 1; i <= m; i++)
            {
                    addedge(u[i] + n,v[i],inf);
                    addedge(v[i] + n,u[i],inf);        
            } 
            Minimum_Cut = 0;
            S = s;
            T = t;
            while (bfs())
            {
                    while (flow = dinic(S,inf)) 
                            Minimum_Cut += flow;
            }
            return Minimum_Cut;
    }
    
    int main() 
    {
            
            while (scanf("%d%d",&n,&m) != EOF)
            {
                    ans = inf;
                    memset(g,false,sizeof(g));
                    for (i = 1; i <= m; i++) 
                    {
                            scanf(" (%d,%d)",&u[i],&v[i]);
                            g[u[i]][v[i]] = g[v[i]][u[i]] = true;
                    }
                    for (s = 0; s < n; s++)        
                    {
                            for (t = 0; t < n; t++)
                            {
                                    if (!g[s][t] && s != t) ans = min(ans,getans(s,t));
                            }
                    }
                    if (n <= 1 || ans == inf) ans = n;
                    printf("%d
    ",ans);
            }
            
            return 0;
        
    }
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  • 原文地址:https://www.cnblogs.com/evenbao/p/9425494.html
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