POJ2195 最小费用流

题目：http://poj.org/problem?id=2195

处理出每个人到每个门的曼哈顿距离，分别建立容量为1费用为曼哈顿距离的边，在源点和每个人人之间建立容量为1费用为0的边，在门和汇点之间建立容量为1费用为0的边，然后跑最小费用流即可。

 #include<cstdio>
 #include<cstring>
 #include<vector>
 #include<queue>
 #include<iostream>
 #include<stdlib.h>
 using namespace std;
 typedef pair<int,int> P;
 const int maxv = ;
 const int inf = 0x3f3f3f3f;
 struct edge{
     int to,cap,cost,rev;
 };
 int V;
 vector<edge> g[maxv];
 int h[maxv];
 int dist[maxv];
 int prevv[maxv],preve[maxv];
 void addedge(int from,int to,int cap,int cost){
     edge t;
     t.to = to;t.cap = cap;t.cost = cost;t.rev = g[to].size();
     g[from].push_back(t);
     t.to = from;t.cap = ;t.cost = -cost;t.rev = g[from].size()-;
     g[to].push_back(t);
 }
 int solve(int s,int t,int f){
     int res = ;
     memset(h,,sizeof(h));
     while(f > ){
         priority_queue<P,vector<P>,greater<P> > que;
         memset(dist,inf,sizeof(dist));
         dist[s] = ;
         que.push(P(,s));
         while(!que.empty()){
             P p = que.top();
             que.pop();
             int v = p.second;
             if(dist[v] < p.first)
                 continue;
             for(int i = ;i<g[v].size();i++){
                 edge &e = g[v][i];
                 if(e.cap> && dist[e.to]>dist[v]+e.cost+h[v]-h[e.to]){
                     dist[e.to] = dist[v]+e.cost+h[v]-h[e.to];
                     prevv[e.to] = v;
                     preve[e.to] = i;
                     que.push(P(dist[e.to],e.to));
                 }
             }
         }
         if(dist[t] == inf){
             return -;
         }
         for(int i = ;i<=t;i++)
             h[i] += dist[i];
         int d = f;
         for(int i = t;i!=s;i = prevv[i]){
             d = min(d,g[prevv[i]][preve[i]].cap);
         }
         f -= d;
         res += d*h[t];
         for(int i = t;i!=s;i = prevv[i]){
             edge &e = g[prevv[i]][preve[i]];
             e.cap -= d;
             g[i][e.rev].cap += d;
         }
     }
     return res;
 }
 int main(){
     int n,m;
     while(scanf("%d%d",&n,&m) && n && m){
         vector<P> mm,hh;
         char ch;
         for(int i = ;i<=n;i++){
             for(int j = ;j<=m;j++){
                 scanf(" %c",&ch);
                 if(ch == 'm')
                     mm.push_back(P(i,j));
                 if(ch == 'H')
                     hh.push_back(P(i,j));
             }
         }
         for(int i = ;i<=*mm.size()+;i++)
             g[i].clear();
         for(int i = ;i<mm.size();i++){
             for(int j = ;j<hh.size();j++){
                 addedge(i+,mm.size()+j+,,abs(mm[i].first-hh[j].first)+abs(mm[i].second-hh[j].second));
             }
         }
         int s = ,t = *mm.size()+;
         for(int i = ;i<mm.size();i++)
             addedge(s,i+,,);
         for(int i = ;i<hh.size();i++)
             addedge(mm.size()+i+,t,,);
         cout << solve(s,t,mm.size()) << endl;
     }
 }