compute eight mat only nts tor between all ans

On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. For each little man, you need to pay a $1 travel fee for every step he moves, until he enters a house. The task is complicated with the restriction that each house can accommodate only one little man.

Your task is to compute the minimum amount of money you need to pay in order to send these n little men into those n different houses. The input is a map of the scenario, a ‘.‘ means an empty space, an ‘H‘ represents a house on that point, and am ‘m‘ indicates there is a little man on that point.

You can think of each point on the grid map as a quite large square, so it can hold n little men at the same time; also, it is okay if a little man steps on a grid with a house without entering that house.

InputThere are one or more test cases in the input. Each case starts with a line giving two integers N and M, where N is the number of rows of the map, and M is the number of columns. The rest of the input will be N lines describing the map. You may assume both N and M are between 2 and 100, inclusive. There will be the same number of ‘H‘s and ‘m‘s on the map; and there will be at most 100 houses. Input will terminate with 0 0 for N and M.
OutputFor each test case, output one line with the single integer, which is the minimum amount, in dollars, you need to pay.
Sample Input

2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0

Sample Output

2
10
28

題意：使得每個m都對應一個H時的最短的路徑

題解：找一個源點，連接所有的m，再找一個匯點，使得所有H指向這個匯點。

然後跑一邊最大費用最小流就ok了

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4
 
 #include <algorithm>
  5 #include <vector>
  6 #include <queue>
  7 #include <cmath>
  8 using namespace std;
  9 typedef long long ll;
 10 const int MAXX=60010;
 11 const int INF=0x3f3f3f3f;
 12 
 13 struct node
 14 {
 15     int st;
 16     int to;
 17     int next;
 18     int cap;
 19     int cost;
 20 }edge[MAXX];
 21 
 22 char mp[102][102];
 23 int head[MAXX],tol;
 24 int pre[MAXX],dis[MAXX];
 25 bool vis[MAXX];
 26 int n,m,p;
 27 
 28 struct node1
 29 {
 30     int x,y;
 31     node1(){}
 32     node1(int a,int b)
 33     {
 34         x=a;
 35         y=b;
 36     }
 37 };
 38 vector<node1> v1,v2;
 39 
 40 void init()
 41 {
 42     tol=0;
 43     memset(head,-1,sizeof(head));
 44     v1.clear();
 45     v2.clear();
 46 }
 47 
 48 void addedge(int u,int v,int cap,int cost)
 49 {
 50     edge[tol].st=u;
 51     edge[tol].to=v;
 52     edge[tol].cap=cap;
 53     edge[tol].cost=cost;
 54     edge[tol].next=head[u];
 55     head[u]=tol++;
 56 
 57     edge[tol].st=v;
 58     edge[tol].to=u;
 59     edge[tol].cap=0;
 60     edge[tol].cost=-cost;
 61     edge[tol].next=head[v];
 62     head[v]=tol++;
 63 }
 64 
 65 bool SPFA(int s,int t)
 66 {
 67     queue<int> q;
 68     memset(dis,INF,sizeof(dis));
 69     memset(vis,0,sizeof(vis));
 70     memset(pre,-1,sizeof(pre));
 71     dis[s]=0;
 72     vis[s]=1;
 73     q.push(s);
 74     while(!q.empty())
 75     { 
 76         int u=q.front(); q.pop();
 77         vis[u]=0;
 78         for(int i=head[u];i!=-1;i=edge[i].next)
 79         {
 80             int to=edge[i].to; 
 81             if(edge[i].cap>0&&dis[to]>dis[u]+edge[i].cost)
 82             {
 83                 dis[to]=dis[u]+edge[i].cost;
 84                 pre[to]=i;
 85                 if(!vis[to])
 86                 {
 87                     vis[to]=1;
 88                     q.push(to);
 89                 }
 90             }
 91         }
 92     }
 93     if(pre[t]==-1)return 0;
 94     return 1;
 95 }
 96 
 97 int minCostMaxFlow(int s,int t)
 98 {
 99     int cost=0;
100     while(SPFA(s,t))
101     { 
102         int minn=INF;
103         for(int i=pre[t];i!=-1;i=pre[edge[i].st])
104             minn=min(minn,edge[i].cap);
105 
106         for(int i=pre[t];i!=-1;i=pre[edge[i].st])
107         {
108             edge[i].cap-=minn;
109             edge[i^1].cap+=minn;
110         }
111         cost+=minn*dis[t];
112     }
113     return cost;
114 }
115 
116 int main()
117 {
118     while(scanf("%d%d",&n,&m)&&m&&n)
119     {
120         getchar();
121         init();
122         for(int i=0;i<n;i++)
123         {
124             scanf("%s",mp[i]);
125             getchar();
126         }
127         for(int i=0;i<n;i++)
128         for(int j=0;j<m;j++)
129         {
130             if(mp[i][j]==‘m‘)
131                 v1.push_back(node1(i,j));
132             if(mp[i][j]==‘H‘)
133                 v2.push_back(node1(i,j));
134 
135         }
136         int l=v1.size();
137         int r=v2.size();
138         for(int i=0;i<v1.size();i++)
139         {
140 
141             node1 N1=v1[i];
142             int x=i+1;
143             addedge(0,x,1,0);
144             for(int j=0;j<v2.size();j++)
145             {
146                 node1 N2=v2[j];
147                 int y=j+l+1;
148                 int D=abs(N1.x-N2.x)+abs(N1.y-N2.y); 
149                 addedge(x,y,1,D);
150                 addedge(y,x,1,D);
151                 if(i==l-1)
152                 addedge(y,l+r+1,1,0);
153             }
154         }
155         int ans=minCostMaxFlow(0,l+1+r);
156         printf("%d\n",ans);
157     }
158     return 0;
159 }

Going Home HDU - 1533（最大費用最小流）