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我也贴个模板,据说是浙大模板#include <stdio.h>
#include <string.h>
#define MAXN 110
#define inf 1000000000
#define _clr(x) memset(x,0xff,sizeof(int)*n)
#define abs(x) ((x)<0?(-(x)):(x))
#define fee(m,h) (abs(m.x-h.x)+abs(m.y-h.y))
typedef struct {
int x,y;
}vex;
//模板开始
//#include <string.h>
//#define MAXN 110
//#define inf 1000000000
//#define _clr(x) memset(x,0xff,sizeof(int)*n)
//二分图最佳匹配,kuhn munkras算法,邻接阵形式,复杂度O(m*m*n)
//返回最佳匹配值,传入二分图大小m,n和邻接阵mat,表示权值
//match1,match2返回一个最佳匹配,未匹配顶点match值为-1
//一定注意m<=n,否则循环无法终止
//最小权匹配可将权值取相反数
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;
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;
}
//模板结束
int main(int argc, char *argv[])
{
int mat[MAXN][MAXN], mat1[MAXN],mat2[MAXN];
int n, m, ans;
char str[MAXN];
vex home[MAXN],men[MAXN];
while (scanf("%d%d\n",&n,&m)==2 && (n||m)){
int p=0;
int q=0;
for(int i=0; i<n; i++){
gets(str);
for(int j=0; j<m; j++){
if(str[j]=='H'){
home[p].x=i;
home[p].y=j;
p++;
}else if(str[j]=='m'){
men[q].x=i;
men[q].y=j;
q++;
}
}
}
for(int i=0; i<q; i++)
for(int j=0; j<p; j++)
mat[i][j] = -fee(men[i], home[j]);//最小用负权
ans = kuhn_munkras(q,p,mat,mat1,mat2);
printf("%d\n",abs(ans));
}
return 0;
}
Memory: 376K Time: 0MS
一次AC,模板V5
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