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。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<vector>
 4 #include<queue>
 5 #include<iostream>
 6 #include<stdlib.h>
 7 using namespace std;
 8 typedef pair<int,int> P;
 9 const int maxv = 222;
10 const int inf = 0x3f3f3f3f;
11 struct edge{
12     int to,cap,cost,rev;
13 };
14 int V;
15 vector<edge> g[maxv];
16 int h[maxv];
17 int dist[maxv];
18 int prevv[maxv],preve[maxv];
19 void addedge(int from,int to,int cap,int cost){
20     edge t;
21     t.to = to;t.cap = cap;t.cost = cost;t.rev = g[to].size();
22     g[from].push_back(t);
23     t.to = from;t.cap = 0;t.cost = -cost;t.rev = g[from].size()-1;
24     g[to].push_back(t);
25 }
26 int solve(int s,int t,int f){
27     int res = 0;
28     memset(h,0,sizeof(h));
29     while(f > 0){
30         priority_queue<P,vector<P>,greater<P> > que;
31         memset(dist,inf,sizeof(dist));
32         dist[s] = 0;
33         que.push(P(0,s));
34         while(!que.empty()){
35             P p = que.top();
36             que.pop();
37             int v = p.second;
38             if(dist[v] < p.first)
39                 continue;
40             for(int i = 0;i<g[v].size();i++){
41                 edge &e = g[v][i];
42                 if(e.cap>0 && dist[e.to]>dist[v]+e.cost+h[v]-h[e.to]){
43                     dist[e.to] = dist[v]+e.cost+h[v]-h[e.to];
44                     prevv[e.to] = v;
45                     preve[e.to] = i;
46                     que.push(P(dist[e.to],e.to));
47                 }
48             }
49         }
50         if(dist[t] == inf){
51             return -1;
52         }
53         for(int i = 0;i<=t;i++)
54             h[i] += dist[i];
55         int d = f;
56         for(int i = t;i!=s;i = prevv[i]){
57             d = min(d,g[prevv[i]][preve[i]].cap);
58         }
59         f -= d;
60         res += d*h[t];
61         for(int i = t;i!=s;i = prevv[i]){
62             edge &e = g[prevv[i]][preve[i]];
63             e.cap -= d;
64             g[i][e.rev].cap += d;
65         }
66     }
67     return res;
68 }
69 int main(){
70     int n,m;
71     while(scanf("%d%d",&n,&m) && n && m){
72         vector<P> mm,hh;
73         char ch;
74         for(int i = 1;i<=n;i++){
75             for(int j = 1;j<=m;j++){
76                 scanf(" %c",&ch);
77                 if(ch == ‘m‘)
78                     mm.push_back(P(i,j));
79                 if(ch == ‘H‘)
80                     hh.push_back(P(i,j));
81             }
82         }
83         for(int i = 0;i<=2*mm.size()+1;i++)
84             g[i].clear();
85         for(int i = 0;i<mm.size();i++){
86             for(int j = 0;j<hh.size();j++){
87                 addedge(i+1,mm.size()+j+1,1,abs(mm[i].first-hh[j].first)+abs(mm[i].second-hh[j].second));
88             }
89         }
90         int s = 0,t = 2*mm.size()+1;
91         for(int i = 0;i<mm.size();i++)
92             addedge(s,i+1,1,0);
93         for(int i = 0;i<hh.size();i++)
94             addedge(mm.size()+i+1,t,1,0);
95         cout << solve(s,t,mm.size()) << endl;
96     }
97 }