POJ 2195 Going Home / HDU 1533（最小费用最大流模板）

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cstdlib>
  5 #include <algorithm>
  6 #include <queue>
  7 
  8 using namespace std;
  9 
 10 struct MCMF{
 11     static const int N = 200 * 200 + 5;
 12     static const int M = 200 * 200 * 4 + 100;
 13     static const int oo = 0x3f3f3f3f;
 14     
 15     int n, m, s, t, tot;
 16     int first[N], next[M];
 17     int u[M], v[M], cap[M], flow[M], cost[M];
 18     int dis[N], a[N], inque[N], pre[N];
 19     
 20     void Clear(){memset(first, -1, sizeof first);tot = 0;}
 21     
 22     void Add(int from, int to, int cp, int flw, int ct){
 23         u[tot] = from; v[tot] = to; cap[tot] = cp; flow[tot] = 0; cost[tot] = ct;
 24         next[tot] = first[u[tot]];
 25         first[u[tot]] = tot; ++ tot;
 26         u[tot] = to; v[tot] = from; cap[tot] = 0; flow[tot] = 0; cost[tot] = -ct;
 27         next[tot] = first[u[tot]];
 28         first[u[tot]] = tot; ++ tot;    
 29     }
 30     
 31     bool bfs(int &flw, int &ct){
 32         for(int i = 0; i <= n + 1; ++ i) dis[i] = oo;
 33         memset(inque, 0, sizeof inque);
 34         dis[s] = 0; pre[s] = 0; a[s] = oo; inque[s] = 1;
 35         
 36         queue <int> q;
 37         q.push(s);
 38         while(!q.empty()){
 39             int now = q.front(); q.pop();
 40             inque[now] = 0;
 41             
 42             for(int i = first[now]; i != -1; i = next[i]){
 43                 if(cap[i] > flow[i] && dis[v[i]] > dis[now] + cost[i]){
 44                     dis[v[i]] = dis[now] + cost[i];
 45                     a[v[i]] = min(a[now], cap[i] - flow[i]);
 46                     pre[v[i]] = i;
 47                     if(!inque[v[i]]){
 48                         inque[v[i]] = 1; q.push(v[i]);
 49                     }
 50                 }
 51             }
 52         }
 53         
 54         if(dis[t] == oo) return false;
 55         flw += a[t];
 56         ct += dis[t] * a[t];
 57         
 58         int now = t;
 59         while(now != s){
 60             flow[pre[now]] += a[t];
 61             flow[pre[now]^1] -= a[t];
 62             now = u[pre[now]];
 63         }
 64         return true; 
 65     }
 66     
 67     int MinCostMaxFlow(int s, int t){
 68         this->s = s;this->t = t;
 69         int flw = 0, ct = 0;
 70         while(bfs(flw, ct));
 71         return ct;
 72     }
 73 }Net;
 74 
 75 struct Position{
 76     int l, r, id;
 77     Position(int _l=0, int _r=0, int _id=0): l(_l), r(_r), id(_id){}
 78 }mm[205], HH[205];
 79 
 80 int ns, ms, cnt1, cnt2, tp1, tp2;
 81 char str[105][105];
 82 
 83 void Solve(){
 84         for(int i = 1; i <= tp1; ++ i){
 85             for(int j = 1; j <= tp2; ++ j){
 86                 int x1 = mm[i].l, y1 = mm[i].r;
 87                 int x2 = HH[j].l, y2 = HH[j].r;
 88                 Net.Add(mm[i].id, HH[j].id, 1, 0, abs(x1-x2) + abs(y1-y2));
 89             }
 90         }
 91 }
 92 
 93 int main(){
 94     
 95     while(scanf("%d%d", &ns, &ms) && ns && ms){
 96         Net.Clear();
 97         cnt1 = cnt2 = 0;
 98         tp1 = tp2 = 0;
 99         for(int i = 1; i <= ns; ++ i)
100             scanf("%s", str[i] + 1);
101         for(int i = 1; i <= ns; ++ i){
102             for(int j = 1; j <= ms; ++ j){
103                 if(str[i][j] == ‘m‘) ++ tp1;
104                 else if (str[i][j] == ‘H‘) ++ tp2;
105             }
106         }
107         Net.n = tp1 + tp2;
108         for(int i = 1; i <= ns; ++ i){
109             for(int j = 1; j <= ms; ++ j){
110                 if(str[i][j] == ‘m‘){
111                     ++ cnt1;
112                     Net.Add(0, cnt1, 1, 0, 0);
113                     mm[cnt1] = (Position){i, j, cnt1};
114                 }
115             }
116         }
117         for(int i = 1; i <= ns; ++ i){
118             for(int j = 1; j <= ms; ++ j){
119                 if(str[i][j] == ‘H‘){
120                     ++ cnt2;
121                     Net.Add(tp1 + cnt2, Net.n + 1, 1, 0, 0);
122                     HH[cnt2] = (Position){i, j, tp1 + cnt2};
123                 }
124             }
125         }
126         Solve();
127         printf("%d\n", Net.MinCostMaxFlow(0, Net.n + 1));
128     }
129     
130     return 0;
131 }