学步园

是为了学习KM算法敲这道题的，虽然是模版题，但是还是看着解题报告才敲出来的，KM还是有点深奥，有些地方没看懂，原理也是似懂非懂.........

先放一放 等以后再细细揣摩

code

#include <set>
#include <map>
#include <ctime>
#include <queue>
#include <cmath>
#include <stack>
#include <limits>
#include <vector>
#include <bitset>
#include <string>
#include <cstdio>
#include <cstring>
#include <fstream>
#include <string.h>
#include <iostream>
#include <algorithm>
#define Si set<int>
#define LL long long
#define pb push_back
#define PS printf(" ")
#define Vi vector<int>
#define LN printf("\n")
#define lson l,m,rt << 1
#define rson m+1,r,rt<<1|1
#define SD(a) scanf("%d",&a)
#define PD(a) printf("%d",a)
#define SET(a,b) memset(a,b,sizeof(a))
#define FF(i,a) for(int i(0);i<(a);i++)
#define FD(i,a) for(int i(a);i>=(1);i--)
#define FOR(i,a,b) for(int i(a);i<=(b);i++)
#define FOD(i,a,b) for(int i(a);i>=(b);i--)
#define readf freopen("input.txt","r",stdin)
#define writef freopen("output.txt","w",stdout)
const int maxn = 102;
const int INF = 0x3fffffff;
const int dx[]={0,1,0,-1};
const int dy[]={1,0,-1,0};
const double pi = acos(-1.0);
const double eps= 1e-7;
using namespace std;
int N,M,nx,ny;
int w[maxn][maxn];
int men[maxn][2],home[maxn][2];
int lx[maxn],ly[maxn];
bool visx[maxn],visy[maxn];
int link[maxn],slack[maxn];
bool find(int x){
    visx[x]=true;
    FOR(y,1,ny){
        if(visy[y]) continue;
        int t=lx[x]+ly[y]-w[x][y];
        if(t==0){
            visy[y]=true;
            if(link[y]==0||find(link[y])){
                link[y]=x;
                return true;
            }
        }else{
            slack[y]=min(slack[y],t);
        }
    }
    return false;
}
int KM(){
    SET(link,0);SET(ly,0);
    FOR(x,1,nx){
        lx[x]=-INF;
        FOR(y,1,ny){
            if(w[x][y]>lx[x]) lx[x]=w[x][y];
        }
    }
    FOR(x,1,nx){
        FOR(y,1,ny){
            slack[y]=INF;
        }
        while(true){
            SET(visx,0);SET(visy,0);
            if(find(x)) break;
            int d=INF;
            FOR(i,1,ny){
                if(!visy[i] && d>slack[i])
                    d=slack[i];
            }
            FOR(i,1,nx){
                if(visx[i]) lx[i]-=d;
            }
            FOR(i,1,ny){
                if(visy[i]) ly[i]+=d;
                else slack[i]-=d;
            }
        }
    }
    int ans=0;
    FOR(i,1,ny){
        ans+=w[link[i]][i];
    }
    return -ans;
}
int main()
{
    char ch;
    while(~scanf("%d%d",&N,&M)&&(N+M)){
        nx=ny=0;
        FOR(i,1,N) FOR(j,1,M){
            scanf(" %c",&ch);
            if(ch=='m'){
                men[++nx][0]=i;
                men[nx][1]=j;
            }
            if(ch=='H'){
                home[++ny][0]=i;
                home[ny][1]=j;
            }
        }
        FOR(i,1,nx) FOR(j,1,ny){
            w[i][j]=-(abs(men[i][0]-home[j][0])+abs(men[i][1]-home[j][1]));
        }
        PD(KM());LN;
    }
    return 0;
}