hdu2195Going Home【最小费用最大流】

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <vector>
  5 #include <cmath>
  6 #include <queue>
  7 using namespace std;
  8 
  9 const int maxn = 1005;
 10 
 11 typedef pair<int, int> PII;
 12 
 13 char a[maxn][maxn];
 14 vector<PII>v[2];
 15 const int INF = 1000000000;
 16 
 17 struct Edge {
 18     int from, to, cap, flow, cost;
 19 };
 20 
 21 struct MCMF
 22 {
 23     int n, m, s, t;
 24     vector<Edge> edges;
 25     vector<int> G[maxn];
 26 
 27     int inq[maxn];
 28     int d[maxn];
 29     int p[maxn];
 30     int a[maxn];
 31 
 32     void init(int n) {
 33         this -> n = n;
 34         for(int i = 0; i < n; i++) G[i].clear();
 35         edges.clear();
 36     }
 37 
 38     void AddEdge(int from, int to, int cap, int cost) {
 39         edges.push_back((Edge) { from, to, cap, 0, cost } );
 40         edges.push_back((Edge) { to, from, 0, 0, -cost } );
 41         m = edges.size();
 42         G[from].push_back(m - 2);
 43         G[to].push_back(m - 1);
 44     }
 45 
 46     bool BellmanFord(int s, int t, int &flow, int &cost) {
 47         for(int i = 0; i < n; i++) d[i] = INF;
 48         memset(inq, 0, sizeof(inq) );
 49         d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
 50         queue<int> Q;
 51         Q.push(s);
 52         while(!Q.empty()) {
 53             int u = Q.front(); Q.pop();
 54             inq[u] = 0;
 55             for(int i = 0; i < G[u].size(); i++) {
 56                 Edge &e = edges[G[u][i]];
 57                 if(e.cap > e.flow && d[e.to] > d[u] + e.cost) {
 58                     d[e.to] = d[u] + e.cost;
 59                     p[e.to] = G[u][i];
 60                     a[e.to] = min(a[u], e.cap - e.flow);
 61                     if(!inq[e.to]) { Q.push(e.to); inq[e.to] = 1; }
 62                 }
 63             }
 64         }
 65 
 66         if(d[t] == INF) return false;
 67         flow += a[t];
 68         cost += d[t] * a[t];
 69         int u = t;
 70         while(u != s) {
 71             edges[p[u]].flow += a[t];
 72             edges[p[u] ^ 1].flow -= a[t];
 73             u = edges[p[u]].from;
 74         }
 75         return true;
 76     }
 77 
 78     int MinCost(int s, int t) {
 79         //this -> s = s; this -> t = t;
 80         int flow = 0, cost = 0;
 81         while(BellmanFord(s, t, flow, cost)){};
 82         return cost;
 83     }
 84 };
 85 
 86 MCMF g;
 87 
 88 int get_dist(PII p1, PII p2) {
 89     return fabs(p1.first - p2.first) + fabs(p1.second - p2.second);
 90 }
 91 
 92 int main() {
 93     int n, m;
 94     //freopen("a.txt","r",stdin);
 95     while(scanf("%d %d",&n, &m)) {
 96         if(n == 0 && m == 0)
 97         break;
 98         //puts("*");
 99         int m_num = 0, H_num = 0;
100         for(int i = 0; i < 2; i++) {
101             v[i].clear();
102         }
103         g.init(n * m);//节点个数  尽量大一点
104         for(int i = 0; i < n; i++) {
105             for(int j = 0; j < m; j++) {
106                 scanf("\n%c",&a[i][j]);//
107                 if(a[i][j] == ‘m‘) {
108                     m_num++;
109                     v[0].push_back(make_pair(i, j));//用vector把所有的人和房间都储存下来
110                 }
111                 else if(a[i][j] == ‘H‘) {
112                     H_num++;
113                     v[1].push_back(make_pair(i, j));
114                 }
115             }
116         }
117         int s = 0; int t = m_num + H_num + 1;
118         //printf("%d %d   %d %d\n",m_num, H_num, v[0].size(), v[1].size());
119         for(int i = 1; i <= m_num; i++) {
120             g.AddEdge(s, i, 1, 0);//源点跟人的容量为1 花费为 0  因为 只能输送一个人
121         }
122         for(int i = m_num + 1; i <= m_num + H_num; i++) {
123             g.AddEdge(i, t, 1, 0);//同理
124         }
125 
126         for(int i = 0; i < v[0].size(); i++) {
127             for(int j = 0; j < v[1].size(); j++) {
128                 g.AddEdge(i + 1, m_num + j + 1, 1, get_dist(v[0][i], v[1][j]));//容量为1 花费为步数
129             }
130         }
131         printf("%d\n",g.MinCost(s, t)) ;
132     }
133     return 0;
134 }