【链接】http://acm.hdu.edu.cn/showproblem.php?pid=1533
【题意】
一个N*M地图上有相同数量的字符H和字符m，m代表一个 人，H代表一个房子。人到房子的花销是它们在图中的曼哈顿距离，问你让所有人回到房子所需要的最小费用（一个房子只能容纳一个人）。
【题解】
费用流;
建立一个超级源点,它和每个房子都有一条边相连,边的容量为1，费用为0；
建立一个超级汇点,他和每个人都有一条边相连,边的容量为1，费用为0；
每个房子和每个人都有一条边,容量为1，费用为它们的曼哈顿距离
这个图的最大流肯定是房子的个数.
则这个时候跑一下最小费用最大流就好.
则这个时候跑一下最小费用最大流就好.
【错的次数】
0
【反思】
第一次写费用流
【代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)
#define ri(x) scanf("%d",&x)
#define rl(x) scanf("%lld",&x)
#define rs(x) scanf("%s",x)
#define oi(x) printf("%d",x)
#define ol(x) printf("%lld",x)
#define oc putchar(' ')
#define os(x) printf(x)
#define all(x) x.begin(),x.end()
#define Open() freopen("F:\rush.txt","r",stdin)
#define Close() ios::sync_with_stdio(0)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 100;
const int NN = 3e4;
const int INF = 0x3f3f3f3f;
struct abc{
int from,en,flow,nex,cost;
};
int n,m,totm,fir[2*N+50],dis[N*2+50],pre[2*N+50],mi[2*N+50],ans;
char s[N+10][N+10];
vector <pii> H,M;
abc bian[NN];
bool inq[2*N+50];
queue <int> dl;
void add(int x,int y,int flow,int cost){
bian[totm].nex = fir[x];
fir[x] = totm;
bian[totm].from = x;
bian[totm].en = y;
bian[totm].cost = cost;
bian[totm].flow = flow;
totm++;
bian[totm].nex = fir[y];
fir[y] = totm;
bian[totm].from = y;
bian[totm].en = x;
bian[totm].cost = -cost;
bian[totm].flow = 0;
totm++;
}
bool spfa(int s,int t){
ms(dis,INF),ms(inq,0),ms(mi,INF);
dis[s] = 0,inq[s] = 1;
dl.push(s);
pre[t] = -1;
while (!dl.empty()){
int x = dl.front();
inq[x] = false;
dl.pop();
for (int i = fir[x];i!=-1;i = bian[i].nex){
int y = bian[i].en;
if (bian[i].flow && dis[y] > dis[x] + bian[i].cost){
dis[y] = dis[x] + bian[i].cost;
mi[y] = min(bian[i].flow,mi[x]);
pre[y] = i;
if (!inq[y]){
inq[y] = true;
dl.push(y);
}
}
}
}
if (pre[t]==-1) return false;
int now = t;
while (now != s){
int temp = pre[now];
bian[temp].flow -= mi[t];
bian[temp^1].flow += mi[t];
now = bian[temp].from;
ans += mi[t]*bian[temp].cost;
}
return true;
}
int main(){
//Open();
//Close();
while (~ri(n)){
ans = 0;
ri(m);
if (n == 0 && m == 0) break;
rep1(i,0,n-1)
rs(s[i]);
H.clear(),M.clear();
rep1(i,0,n-1)
rep1(j,0,m-1)
if (s[i][j]=='H')
H.pb(mp(i,j));
else if (s[i][j]=='m')
M.pb(mp(i,j));
totm = 0;
ms(fir,255);
int len1 = H.size(),len2 = M.size();
rep1(i,0,len1-1) add(0,i+1,1,0);
rep1(i,0,len2-1) add(i+1+len1,len1+len2+1,1,0);
rep1(i,0,len1-1)
rep1(j,0,len2-1)
add(i+1,len1+j+1,1,abs(M[j].fi-H[i].fi)+abs(M[j].se-H[i].se));
while (spfa(0,len1+len2+1));
oi(ans);puts("");
}
return 0;
}