POJ 2195

#include<iostream>//by Chengdacaizi
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<math.h>
#define MAXN 105
#include <string>
#define inf 1000000000
#define _clr(x) memset(x,0xff,sizeof(int)*n)
using namespace std;

int _m[MAXN][MAXN];
vector<int> man;
vector<int> hou;
int match1[MAXN];
int match2[MAXN];
int kuhn_munkras(int m,int n,int mat[][MAXN],int* match1,int* match2);

int main()
{
    //freopen("acm.acm","r",stdin);
    int n;
    int m;
    int i;
    int j;
    char c;
    int tem1;
    int tem2;
    int temj1;
    int temj2;
    while(cin>>n>>m)
    {
        memset(_m,0,sizeof(_m));
        if(!n&&!m)
            break;
        for(i = 0; i < n; ++ i)
        {
            for(j = 0; j < m; ++ j)
            {
                cin>>c;
                if(c == 'm')
                    man.push_back(i*m+j);
                else if(c == 'H')
                    hou.push_back(i*m+j);
            }
        }
        for(i = 0; i < man.size(); ++ i)
        {
            for(j = 0; j < hou.size(); ++ j)
            {
                tem1 = man[i]/m;
                tem2 = man[i]%m;
                temj1 = hou[j]/m;
                temj2 = hou[j]%m;
                _m[i][j] = (-1)*(abs(tem1 - temj1) + abs(tem2 - temj2));
            }
        }
        
        m = man.size();
        n = hou.size();
        if(m>n)
            iter_swap(&m,&n);
        cout<<-kuhn_munkras(m,n,_m,match1,match2)<<endl;
        man.clear();
        hou.clear();
    }
}





int kuhn_munkras(int m,int n,int mat[][MAXN],int* match1,int* match2){
    int s[MAXN],t[MAXN],l1[MAXN],l2[MAXN],p,q,ret=0,i,j,k;//最佳匹配
    for (i=0;i<m;i++)
        for (l1[i]=-inf,j=0;j<n;j++)
            l1[i]=mat[i][j]>l1[i]?mat[i][j]:l1[i];
    for (i=0;i<n;l2[i++]=0);
    for (_clr(match1),_clr(match2),i=0;i<m;i++){
        for (_clr(t),s[p=q=0]=i;p<=q&&match1[i]<0;p++)
            for (k=s[p],j=0;j<n&&match1[i]<0;j++)
                if (l1[k]+l2[j]==mat[k][j]&&t[j]<0){
                    s[++q]=match2[j],t[j]=k;
                    if (s[q]<0)
                        for (p=j;p>=0;j=p)
                            match2[j]=k=t[j],p=match1[k],match1[k]=j;
                }
        if (match1[i]<0){
            for (i--,p=inf,k=0;k<=q;k++)
                for (j=0;j<n;j++)
                    if (t[j]<0&&l1[s[k]]+l2[j]-mat[s[k]][j]<p)
                        p=l1[s[k]]+l2[j]-mat[s[k]][j];
            for (j=0;j<n;l2[j]+=t[j]<0?0:p,j++);
            for (k=0;k<=q;l1[s[k++]]-=p);
        }
    }
    for (i=0;i<m;i++)
        ret+=mat[i][match1[i]];
    return ret;
}

 

关注我的公众号，当然，如果你对Java, Scala, Python等技术经验，以及编程日记，感兴趣的话。 

技术网站地址： vmfor.com