二分图KM算法 POJ 2195

暴力（n^3）

#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAX 105
#define INF 9999999
struct House { int r, c; } house[MAX];
struct Man { int r, c; } man[MAX];
int H, M, n, m;
int A[MAX], B[MAX];
int visA[MAX], visB[MAX];
int match[MAX], slack[MAX], map[MAX][MAX];

bool find_path ( int i )
{
	visA[i] = true;
	for ( int j = 0; j < H; j++ )
	{
		if ( !visB[j] && A[i] + B[j] == map[i][j] )
		{
			visB[j] = true;
			if (match[j] == -1 || find_path(match[j]))
			{
				match[j] = i;
				return true;
			}
		}
		else if ( A[i] + B[j] > map[i][j] ) //j属于B，且不在交错路径中
			slack[j] = min(slack[j], A[i]+B[j]-map[i][j]);
	}
	return false;
}

void KM ()
{
    int i, j, d;
    memset(A,0,sizeof(A));
    memset(B,0,sizeof(B));
    memset(match,-1,sizeof(match));
    for ( i = 0; i < M; i++ )
        for ( j = 0; j < H; j++ )
            A[i] = max (map[i][j], A[i]);
    for ( i = 0; i < M; i++ )
    {
        for ( j = 0; j < H; j++ )
            slack[j] = INF;
        while ( 1 )
        {
            memset(visA,0,sizeof(visA));
            memset(visB,0,sizeof(visB));
            if ( find_path ( i ) ) break; //从i点出发找到交错路径则跳出循环
            for ( d = INF, j = 0; j < H; j++ ) //取最小的slack[j]
                if (!visB[j] && d > slack[j]) d = slack[j];
            for ( j = 0; j < M; j++ ) //集合A中位于交错路径上的-d
                if ( visA[j] ) A[j] -= d;
            for ( j = 0; j < H; j++ ) //集合B中位于交错路径上的+d
                if ( visB[j] ) B[j] += d;
                else slack[j] -= d; //注意修改不在交错路径上的slack[j]
        }
    }
}

int main()
{
    char s[MAX];
	int i, j, res;
	while ( scanf("%d%d",&n,&m) )
	{
	    if ( !m && !n ) break;
		H = M = res = 0;
		for ( i = 0; i < n; i++ )
		{
			scanf("%s",s);
			for ( j = 0; j < m; j++ )
			{
				if ( s[j] == 'H' )
                    house[H].r = i, house[H++].c = j;
				else if ( s[j] == 'm' )
					man[M].r = i, man[M++].c = j;
			}
		}
		for ( i = 0; i < M; i++ ) //求最小带权匹配可以将权值改为负数
			for ( j = 0; j < H; j++ )
				map[i][j] = -(abs(man[i].r-house[j].r) + abs(man[i].c-house[j].c));
        KM();
		for ( j = 0; j < H; j++ )
			res -= map[match[j]][j];
		printf("%d\n",res);
	}
	return 0;
}

KM算法

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
const int maxn = 105;
const int INF = 1<<29;
int w[maxn][maxn];
int n;
int ly[maxn],lx[maxn];//顶标
bool S[maxn],T[maxn]; //左/右第i个点是否已经标记
int left[maxn];       //匹配为右边第i个点的编号
int H, M;
int R,C;
bool match(int i){
   S[i]=1;
   for(int j=1;j<=n;j++)
       if(lx[i]+ly[j]==w[i][j]&&!T[j]){
          T[j]=1;
          if(!left[j]||match(left[j])){
            left[j]=i;
            return true;
          }
       }
   return false;
}

void update(){
    int a=INF;
    for(int i=1;i<=n;i++){
        if(S[i])
        for(int j=1;j<=n;j++){
            if(!T[j])
                a=min(a,lx[i]+ly[j]-w[i][j]);
        }
    }

    for(int i=1;i<=n;i++){
        if(S[i]) lx[i]-=a;
        if(T[i]) ly[i]+=a;
    }
}

void KM(){
    for(int i=1;i<=n;i++){
        left[i]=lx[i]=ly[i]=0;
        for(int j=1; j<=n;j++){
            lx[i]=max(lx[i],w[i][j]);
        }
    }
    for(int i=1;i<=n;i++){
        while(1){
                //printf("1");
            memset(T,0,sizeof(T));
            memset(S,0,sizeof(S));
            if(match(i)) break;
            else update();
        }
    }
}
struct House { int r, c; } house[maxn];
struct Man { int r, c; } man[maxn];

int main()
{
    char s[maxn];
	int i, j;
	while ( scanf("%d %d",&R,&C) )
	{
	    if ( !R && !C ) break;
		H = M = 1;
		for ( i = 1; i <=R; i++ )
		{
			scanf("%s",s+1);
			for ( j = 1; j <=C; j++ )
			{
				if ( s[j] == 'H' )
                    house[H].r = i, house[H++].c = j;
				else if ( s[j] == 'm' )
					man[M].r = i, man[M++].c = j;
			}
		}
		for ( i = 1; i <= M-1; i++ ) //求最小带权匹配可以将权值改为负数
			for ( j = 1; j <= H-1; j++ )
				w[i][j] = -(abs(man[i].r-house[j].r) + abs(man[i].c-house[j].c));
        n=H-1;
        KM();
        int res=0;
		for ( j = 1; j <=H-1; j++ )
			res -= w[left[j]][j];
		printf("%d\n",res);
	}
	return 0;
}