嗯，就是说一个n*m的地图
然后上面有相等数量的小人和房子，小人每次可以上下左右地走到相邻点，然后每个小人要走到一所房子里，每个房子也只能装一个小人
然后小人走到房子的花费就是小人走的步数，一个点上可以有多个小人，一个小人也可以走到一个房子的点上但是不进入这个 房子
嗯，这样就是很裸的费用流，建一个超级源点，连接所有小人，容量为1，费用为0，建一个超级汇点，连接所有房子，容量为1，费用为0，
然后每个小人与每个房子，建边，容量为1，费用为他们的曼哈顿距离
然后从超级源点往超级汇点跑费用流
#include <string>
#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <vector>
#include <fstream>
#include <queue>
using namespace std;
#define ll long long
#define INF 0x3f3f3f3f
#define maxn 10005
#define maxm 100005
//最小费用最大流，求最大费用只需要取相反数，结果取相反数即可。
//点的总数为 N，点的编号[0,N-1]
struct Edge
{
int to, next, cap, flow, cost;
}edge[maxm];
int head[maxn], tol;
int pre[maxn], dist[maxn];
bool vis[maxn];
int N;//节点总个数，节点编号从0~N-1
void init(int n)
{
N = n;
tol = 0;
memset(head, -1, sizeof(int)*maxn);
}
void addedge(int u, int v, int cap, int cost)
{
edge[tol].to = v;
edge[tol].cap = cap;
edge[tol].cost = cost;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = 0;
edge[tol].cost = -cost;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
bool spfa(int s, int t)
{
queue<int>q;
memset(dist, 0x3f, sizeof(int)*maxn);
memset(vis, 0, sizeof(vis));
memset(pre, -1, sizeof(int)*maxn);
dist[s] = 0;
vis[s] = true;
q.push(s);
while (!q.empty())
{
int u = q.front();
q.pop();
vis[u] = false;
for (int i = head[u]; i != -1; i = edge[i].next)
{
int v = edge[i].to;
if (edge[i].cap > edge[i].flow &&
dist[v] > dist[u] + edge[i].cost)
{
dist[v] = dist[u] + edge[i].cost;
pre[v] = i;
if (!vis[v])
{
vis[v] = true;
q.push(v);
}
}
}
}
if (pre[t] == -1)return false;
else return true;
}
//返回的是最大流，cost存的是最小费用
int minCostMaxflow(int s, int t, int &cost)
{
int flow = 0;
cost = 0;
while (spfa(s, t))
{
int Min = INF;
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
{
if (Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
}
for (int i = pre[t]; i != -1; i = pre[edge[i ^ 1].to])
{
edge[i].flow += Min;
edge[i ^ 1].flow -= Min;
cost += edge[i].cost * Min;
}
flow += Min;
}
return flow;
}
int n, m;
pair<int, int> men[105], house[105];
char s0[105];
int main()
{
//freopen("input.txt","r",stdin);
//freopen("output.txt","w",stdout);
//ios::sync_with_stdio(false);
//cin.tie(0); cout.tie(0);
//ifstream in;
//in.open("input.txt", ios::in);
while (scanf("%d%d", &n, &m) != EOF)
{
if (n == 0 && m == 0)
break;
int nm = 0, nh = 0;
for (int i = 0; i < n; ++i)
{
scanf("%s", s0);
for (int j = 0; j < m; ++j)
{
if (s0[j] == 'm')
men[nm++] = make_pair(i, j);
if (s0[j] == 'H')
house[nh++] = make_pair(i, j);
}
}
init(nm + nh + 2);
/*for (int i = 0; i < nm; ++i)
printf("%d %d ", men[i].first, men[i].second);
printf("\n");
for (int i = 0; i < nh; ++i)
printf("%d %d ", house[i].first, house[i].second);
printf("\n");*/
for (int i = 0; i < nm; ++i)
addedge(0, i + 1, 1, 0);
for (int i = 0; i < nh; ++i)
addedge(nm + i + 1, nm + nh + 1, 1, 0);
for (int i = 0; i < nm; ++i)
{
for (int j = 0; j < nh; ++j)
addedge(i + 1, nm + j + 1, 1, abs(men[i].first - house[j].first) + abs(men[i].second - house[j].second));
}
int ans = 0;
minCostMaxflow(0, nm + nh + 1, ans);
printf("%d\n", ans);
}
//while (1);
return 0;
}