HDU1533Going Home（KM匹配之最小值匹配）

题意：求最小花费。ＫＭ通常是来求最大完美匹配，这里只需要把权重变为负数。最后再变回来即可

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<string>
#include<cstring>
#include<stack>
#include<queue>
#include<vector>
#include<cstdlib>
#define lson (rt<<1),L,M
#define rson (rt<<1|1),M+1,R
#define M ((L+R)>>1)
#define cl(a,b) memset(a,b,sizeof(a));
#define LL long long
#define P pair<int,int>
#define X first
#define Y second
#define pb push_back
#define fread(zcc)  freopen(zcc,"r",stdin)
#define fwrite(zcc) freopen(zcc,"w",stdout)
using namespace std;
const int maxn=305;
const int inf=999999;

int w[maxn][maxn];
int linker[maxn],lx[maxn],ly[maxn],slack[maxn];
bool visx[maxn],visy[maxn];
int nx,ny;

bool dfs(int x){
    visx[x]=true;
    for(int y=1;y<=ny;y++){
        if(visy[y])continue;
        int tmp=lx[x]+ly[y]-w[x][y];
        if(tmp==0){
            visy[y]=true;
            if(linker[y]==-1||dfs(linker[y])){
                linker[y]=x;
                return true;
            }
        }else if(slack[y]>tmp){
            slack[y]=tmp;
        }
    }
    return false;
}
int km(){
    cl(linker,-1);
    cl(ly,0);
    for(int i=1;i<=nx;i++){
        lx[i]=-inf;
        for(int j=1;j<=ny;j++)if(w[i][j]>lx[i]){
            lx[i]=w[i][j];
        }
    }
    for(int x=1;x<=nx;x++){
        fill(slack,slack+ny+1,inf);
        while(true){
            cl(visx,false);
            cl(visy,false);
            if(dfs(x))break;
            int d=inf;
            for(int i=1;i<=ny;i++)if(!visy[i]&&d>slack[i]){
                d=slack[i];
            }
            for(int i=1;i<=nx;i++)if(visx[i]){
                lx[i]-=d;
            }
            for(int i=1;i<=ny;i++)if(visy[i])ly[i]+=d;
            else slack[i]-=d;
        }
    }
    int ans=0;
    for(int i=1;i<=ny;i++)if(linker[i]!=-1){
        ans+=w[linker[i]][i];
    }
    return ans;
}
char a[maxn][maxn];
P h[maxn],mm[maxn];
inline int abs(int s){return s<0?-s:s;}
int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)&&n+m){
        int num1=1,num2=1;
        for(int i=1;i<=n;i++){
            scanf("%s",a[i]);
            for(int j=0;j<m;j++){
                if(a[i][j]==‘H‘){
                    h[num1].X=i;
                    h[num1++].Y=j;
                }
                if(a[i][j]==‘m‘){
                    mm[num2].X=i;
                    mm[num2++].Y=j;
                }
            }
        }
        //printf("===%d   %d\n",num1,num2);
        for(int i=1;i<num1;i++){
            for(int j=1;j<num2;j++){
                w[i][j]=abs(h[i].X-mm[j].X)+abs(h[i].Y-mm[j].Y);
                w[i][j]=-w[i][j];
            }
        }
        nx=ny=num2-1;
        printf("%d\n",-km());
    }
    return 0;
}