Description
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.
Input
There 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.
Output
For each test case, output one line with the single integer, which is the minimum amount, in dollars, you need to pay.
Sample Input
代码语言：javascript复制2 2
.m
H.
5 5
HH..m
.....
.....
.....
mm..H
7 8
...H....
...H....
...H....
mmmHmmmm
...H....
...H....
...H....
0 0Sample Output
代码语言：javascript复制2
10
28就是普通的费用流问题， 源点s编号0, 人编号1到n, 房子编号n 1到n n, 汇点编号t， 源点s到每个人i有边(s, i, 1,0）， 每个人i到每个房子j有边(i, j, 1, i人到j房的开销)， 每个房子j到汇点t有边(j, t, 1, 0)。就成了最基本的费用流。
代码语言：javascript复制#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>
#include<cmath>
#define INF 1e9
using namespace std;
const int maxn=200 5;
struct Edge
{
int from,to,cap,flow,cost;
Edge(){}
Edge(int f,int t,int c,int fl,int co):from(f),to(t),cap(c),flow(fl),cost(co){}
};
struct MCMF
{
int n,m,s,t;
vector<Edge> edges;
vector<int> G[maxn];
bool inq[maxn];
int d[maxn];
int p[maxn];
int a[maxn];
void init(int n,int s,int t)
{
this->n=n, this->s=s, this->t=t;
edges.clear();
for(int i=0;i<n; i) G[i].clear();
}
void AddEdge(int from,int to,int cap,int cost)
{
edges.push_back(Edge(from,to,cap,0,cost));
edges.push_back(Edge(to,from,0,0,-cost));
m=edges.size();
G[from].push_back(m-2);
G[to].push_back(m-1);
}
bool BellmanFord(int &flow,int &cost)
{
for(int i=0;i<n; i) d[i]=INF;
queue<int> Q;
memset(inq,0,sizeof(inq));
d[s]=0, Q.push(s), a[s]=INF, p[s]=0, inq[s]=true;
while(!Q.empty())
{
int u=Q.front(); Q.pop();
inq[u]=false;
for(int i=0;i<G[u].size(); i)
{
Edge &e=edges[G[u][i]];
if(e.cap>e.flow && d[e.to]>d[u] e.cost)
{
d[e.to]=d[u] e.cost;
p[e.to]=G[u][i];
a[e.to]=min(a[u],e.cap-e.flow);
if(!inq[e.to]){ Q.push(e.to); inq[e.to]=true; }
}
}
}
if(d[t]==INF) return false;
flow =a[t];
cost =a[t]*d[t];
int u=t;
while(u!=s)
{
edges[p[u]].flow =a[t];
edges[p[u]^1].flow -=a[t];
u=edges[p[u]].from;
}
return true;
}
int solve()
{
int flow=0,cost=0;
while(BellmanFord(flow,cost));
return cost;
}
}MM;
struct Node
{
int x,y;
Node(){}
Node(int x,int y):x(x),y(y){}
int get_dist(Node& b)
{
return abs(x-b.x) abs(y-b.y);
}
}node1[maxn],node2[maxn];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)==2 && n)
{
int num1=0,num2=0;//记录人数和房子数
for(int i=1;i<=n; i)
for(int j=1;j<=m; j)
{
char ch;
scanf(" %c",&ch);
if(ch=='m') node1[num1 ]=Node(i,j);
else if(ch=='H') node2[num2 ]=Node(i,j);
}
int src=0,dst=num1*2 1;
MM.init(num1*2 2,src,dst);
for(int i=1;i<=num1; i)
{
MM.AddEdge(src,i,1,0);
MM.AddEdge(num1 i,dst,1,0);
for(int j=1;j<=num1; j)
{
MM.AddEdge(i,num1 j,1,node1[i-1].get_dist(node2[j-1]));
}
}
printf("%dn",MM.solve());
}
return 0;
}
