HDU-2195-Going Home

最新推荐文章于 2020-01-27 10:07:07 发布

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#arrays #path #matrix #n2 #struct #算法

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本文介绍了一种使用最大带权匹配算法(KM算法)解决HDOJ平台上的HDU-2195 GoingHome问题的方法。通过构建二分图并将所有H节点作为X集合，所有m节点作为Y集合，实现对问题的有效求解。

HDU-2195-Going Home

http://acm.hdu.edu.cn/showproblem.php?pid=1533

之前用最小费用最大流做的，今天看了最大带权匹配的KM算法，套用了模板来做这题，将所有的H作为X集合，所有的m作为Y集合，构造二分图求最大带权匹配

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
using namespace std;
const int MAXN=5000;
const int INF=0x7fffffff;
int nx, ny, w[MAXN][MAXN], lx[MAXN], ly[MAXN];
int fx[MAXN], fy[MAXN], matx[MAXN], maty[MAXN];
char map[MAXN][MAXN];
int n1,n2;
struct node
{
	int x;
	int y;
}h[MAXN],man[MAXN];
//模板
//  Name: PerfectMatch by Kuhn_Munkras O(n^3)
//  Description: w is the adjacency matrix, nx,ny are the size of x and y,
//  lx, ly are the lables of x and y, fx[i], fy[i] is used for marking
//  whether the i-th node is visited,  matx[x] means x match matx[x],
//  maty[y] means y match maty[y], actually, matx[x] is useless,
//  all the arrays are start at 1
int path(int u)
{  
    fx[u] = 1;
    for (int v = 1; v <= ny; v++) 
    if (lx[u] + ly[v] == w[u][v] && fy[v] < 0)
    {   
            fy[v] = 1;
            if (maty[v] < 0 || path(maty[v]))
            {    
                matx[u] = v;    
                maty[v] = u;    
                return 1;   
            }
     }
     return 0;
}
int km()
{  
    int i,j,k,ret = 0;
    memset(ly, 0, sizeof(ly));
    for (i = 1; i <= nx; i++)
    {  
        lx[i] = -INF;
        for (j = 1; j <= ny; j++) 
        if (w[i][j] > lx[i]) lx[i] = w[i][j];
    }
    memset(matx, -1, sizeof(matx));
    memset(maty, -1, sizeof(maty));
    for (i = 1; i <= nx; i++)
    {   
        memset(fx, -1, sizeof(fx));
        memset(fy, -1, sizeof(fy));
        if (!path(i))
        {  
            i--;   
            int p = INF;
            for (k = 1; k <= nx; k++)
			{
                if (fx[k] > 0)
                for (j = 1; j <= ny; j++) 
                if (fy[j] < 0 && lx[k] + ly[j] - w[k][j] < p)
                p=lx[k]+ly[j]-w[k][j];
            }
            for (j = 1; j <= ny; j++) 
            ly[j] += fy[j]<0 ? 0 : p;
            for (j = 1; j <= nx; j++) 
            lx[j] -= fx[j]<0 ? 0 : p;
        }
    }
    for (i = 1; i <= ny; i++)
	ret += w[maty[i]][i];
    return ret;
}
 void init()
{
	int i,j,v;
	char c;
	nx=ny=0;
	for(i=0;i<n1;i++)
	{
		for(j=0;j<n2;j++)
		{
			scanf("%c",&c);
			if(c=='H')
			{
				h[++nx].x=i;
				h[nx].y=j;
			}
			else if(c=='m')
			{
				man[++ny].x=i;
				man[ny].y=j;
			}
		}
		getchar();
	}
	for(i=1;i<=nx;i++)
	for(j=1;j<=ny;j++)
	w[i][j]=-99999999;
	for(i=1;i<=nx;i++)
	for(j=1;j<=ny;j++)
	w[i][j]=-(abs(h[i].x-man[j].x)+abs(h[i].y-man[j].y));
 }
int main()
{
	while(scanf("%d %d",&n1,&n2),n1||n2)
	{
		getchar();
		init();
		printf("%d\n",-km());
	}
	return 0;
}