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 = 222;
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 = 0;t.cost = -cost;t.rev = g[from].size()-1;
    g[to].push_back(t); 
}
int solve(int s,int t,int f){
    int res = 0;
    memset(h,0,sizeof(h));
    while(f > 0){
        priority_queue<P,vector<P>,greater<P> > que;
        memset(dist,inf,sizeof(dist));
        dist[s] = 0;
        que.push(P(0,s));
        while(!que.empty()){
            P p = que.top();
            que.pop();
            int v = p.second;
            if(dist[v] < p.first)
                continue;
            for(int i = 0;i<g[v].size();i++){
                edge &e = g[v][i];
                if(e.cap>0 && 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 -1;
        }
        for(int i = 0;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 = 1;i<=n;i++){
            for(int j = 1;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 = 0;i<=2*mm.size()+1;i++)
            g[i].clear();
        for(int i = 0;i<mm.size();i++){
            for(int j = 0;j<hh.size();j++){
                addedge(i+1,mm.size()+j+1,1,abs(mm[i].first-hh[j].first)+abs(mm[i].second-hh[j].second));
            }
        }
        int s = 0,t = 2*mm.size()+1;
        for(int i = 0;i<mm.size();i++)
            addedge(s,i+1,1,0);
        for(int i = 0;i<hh.size();i++)
            addedge(mm.size()+i+1,t,1,0);
        cout << solve(s,t,mm.size()) << endl;
    } 
}