poj2195Going Home(最小费用最大流）

 

http://poj.org/problem?id=2195

模板题 超级源点-》men->house->超级汇点 坐标差值为两点的cost值 cap为1

View Code 
  1 #include <iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #include<queue>
  6 #include<stdlib.h>
  7 #define INF 0xfffffff
  8 using namespace std;
  9 struct node
 10 {
 11     int u,v,next,cost,cap;
 12 }edge[210000];
 13 int men[1100][2],house[1100][2],head[1100],t,dist[1100],vis[1100],pp[1100];
 14 void init()
 15 {
 16     t = 0;
 17     memset(head,-1,sizeof(head));
 18 }
 19 void add(int u,int v,int c,int p)
 20 {
 21     edge[t].u = u;
 22     edge[t].v = v;
 23     edge[t].cap = c;
 24     edge[t].cost = p;
 25     edge[t].next = head[u];
 26     head[u] = t++;
 27     edge[t].v = u;
 28     edge[t].u = v;
 29     edge[t].cap = 0;
 30     edge[t].cost = -p;
 31     edge[t].next = head[v];
 32     head[v] = t++;
 33 }
 34 int spfa(int s,int en,int n)
 35 {
 36     int u,v,i;
 37     queue<int>q;
 38     memset(vis,0,sizeof(vis));
 39     memset(pp,-1,sizeof(pp));
 40     for(i = 0 ; i <= n ; i++)
 41     dist[i] = INF;
 42     dist[s] = 0;
 43     vis[s] = 1;
 44     q.push(s);
 45     while(!q.empty())
 46     {
 47         u = q.front();
 48         q.pop();
 49         vis[u] = 0;
 50         for(i = head[u] ; i != -1 ; i = edge[i].next)
 51         {
 52             v = edge[i].v;
 53             if(edge[i].cap&&dist[v]>dist[u]+edge[i].cost)
 54             {
 55                 dist[v] = dist[u]+edge[i].cost;
 56                 pp[v] = i;
 57                 if(!vis[v])
 58                 {
 59                     vis[v] = 1;
 60                     q.push(v);
 61                 }
 62             }
 63         }
 64     }
 65     if(dist[en]==INF)
 66     return 0;
 67     return 1;
 68 }
 69 int mcmf(int s,int en,int n)
 70 {
 71     int i,mc=0,minf;
 72     while(spfa(s,en,n))
 73     {
 74         minf = INF+1;
 75         for(i = pp[en] ; i != -1 ; i = pp[edge[i].u])
 76         if(minf>edge[i].cap)
 77         minf = edge[i].cap;
 78         for(i = pp[en]; i!=-1 ;i = pp[edge[i].u])
 79         {
 80             edge[i].cap-=minf;
 81             edge[i^1].cap+=minf;
 82         }
 83         mc+=minf*dist[en];
 84     }
 85     return mc;
 86 }
 87 int main()
 88 {
 89     int i,j,k,n,m,g,o;
 90     char c;
 91     while(cin>>n>>m)
 92     {
 93         if(n==0&&m==0)
 94             break;
 95         o=0;g=0;
 96         init();
 97         for(i = 1; i <= n ; i++)
 98         {
 99             getchar();
100             for(j = 1; j <= m ; j++)
101             {
102                 scanf("%c",&c);
103                 if(c=='m')
104                 {
105                     g++;
106                     men[g][0] = i;
107                     men[g][1] = j;
108                 }
109                 if(c=='H')
110                 {
111                     o++;
112                     house[o][0] = i;
113                     house[o][1] = j;
114                 }
115             }
116         }
117         int s = 0 ,en = g+o+1;
118         for( i = 1 ; i <= g ; i++)
119         add(s,i,1,0);
120         for(i = 1 ; i <= o ;i++)
121         add(i+g,en,1,0);
122         for(i = 1 ; i <= g ; i++)
123             for(j = 1; j <= o ; j++)
124             {
125                 int x = abs(men[i][0]-house[j][0])+abs(men[i][1]-house[j][1]);
126                 add(i,g+j,1,x);
127             }
128         cout<<mcmf(s,en,g+o+1)<<endl;
129     }
130     return 0;
131 }

 

 

 

posted @ 2013-03-23 00:46  _雨  阅读(236)  评论(0)  编辑  收藏  举报