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  • [POJ 2135] Farm Tour

    [题目链接]

             http://poj.org/problem?id=2135

    [算法]

            费用流

    [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;
    #define MAXN 5010
    #define MAXM 20010
    const int inf = 2e9;
    
    int i,n,m,u,v,w,S,T,ans,tot;
    int head[MAXN],pre[MAXN],incf[MAXN],dist[MAXN];
    
    struct edge
    {
            int to,w,cost,nxt;
    } e[MAXM << 1];
    
    inline void addedge(int u,int v,int w,int cost)
    {
            tot++;
            e[tot] = (edge){v,w,cost,head[u]};
            head[u] = tot;
            tot++;
            e[tot] = (edge){u,0,-cost,head[v]};
            head[v] = tot;
    }
    inline bool spfa()
    {
            int i,l,r,u,v,w,cost;
            static int q[MAXN];
            static bool inq[MAXN];
            for (i = 1; i <= T; i++) dist[i] = inf;
            memset(inq,false,sizeof(inq));
            q[l = r = 1] = S;
            inq[S] = true;
            pre[S] = 0;
            dist[S] = 0;
            incf[S] = inf;
            while (l <= r)
            {
                    u = q[l];
                    l++;
                    inq[u] = false;
                    for (i = head[u]; i; i = e[i].nxt)
                    {
                            v = e[i].to;
                            w = e[i].w;
                            cost = e[i].cost;
                            if (w && dist[u] + cost < dist[v])
                            {
                                    dist[v] = dist[u] + cost;
                                    incf[v] = min(incf[u],w);
                                    pre[v] = i;
                                    if (!inq[v])
                                    {
                                            inq[v] = true;
                                            q[++r] = v;
                                    }        
                            }        
                    }        
            }        
            if (dist[T] != inf) return true;
            else return false;
    } 
    inline void update()
    {
            int x = T,pos;
            while (x != S)
            {
                    pos = pre[x];
                    e[pos].w -= incf[T];
                    e[pos ^ 1].w += incf[T];
                    x = e[pos ^ 1].to; 
            }
            ans += dist[T] * incf[T];
    }
    int main() 
    {
            
            scanf("%d%d",&n,&m);
            tot = 1;
            for (i = 1; i <= m; i++)
            {
                    scanf("%d%d%d",&u,&v,&w);
                    addedge(u,v,1,w);
                    addedge(v,u,1,w);
            }
            S = n + 1;
            T = n + 2;
            addedge(S,1,2,0);
            addedge(n,T,2,0);
            while (spfa()) update();
            printf("%d
    ",ans);
            
            return 0;
        
    }
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  • 原文地址:https://www.cnblogs.com/evenbao/p/9426229.html
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