2060.txt

北大ACM题目例程 详细的解答过程 程序实现 算法分析

#include <memory.h>
#include <stdio.h>
#include <math.h>
#define null 0
#define y1 yy1
const int size = 510;
//size must be bigger than n and m.
//n小于m时效率高
int maxmatch( int n, int m, bool w[][size], int *p)
{

	int p_n[size];
	int p_m[size];
	bool sign[size];
	int q[size],from[size],s,t;
	int i,j,link,now,h;
	memset( p_n, -1, sizeof(int)*n );
	memset( p_m, -1, sizeof(int)*m );
	for(i=0;i<n;i++)
	if(p_n[i]==-1)
	{
		memset( sign, 0, sizeof(bool)*m );
                s=1;link=-1;
		from[0]=-1;
		q[0]=size-1;
		p_m[size-1]=i;
		for(t=0;t<s;t++)
		{
		   now=q[t];
		   for(j=0;j<m;j++)
		   {
			   if( w[p_m[now]][j] != null && sign[j]==0 )
			   {
					sign[j]=1;
					q[s]=j;
					from[s++]=t;

					if(p_m[j]==-1)
					{
						link=s-1;
						break;
					}
			   }
		   }
		   if(j<m)break;
		}
		if(t<s)
		{
			 while(from[link]!=-1)
			 {
					h=from[link];
					p_m[q[link]]=p_m[q[h]];
					p_n[p_m[q[h]]]=q[link];
					link=h;
			 }
		}

		//无完备匹配
		//else return 0;
	}
	int an;
	for(i=0,an=0;i<n;i++)
	{
		if(p)p[i]=p_n[i];
		if(p_n[i]>=0)an++;
	}
	return an;
}
bool e[size][size];
int begin[size], end[size];
int x1[size], y1[size], x2[size], y2[size], n;
inline int dis( int x1, int y1, int x2, int y2 )
{
	return abs( x1 - x2 ) + abs( y1 - y2 );
}
int main()
{
	int cas, n, i, j, a, b;	
	scanf( "%d", &cas );
	while( cas-- )
	{
		memset( e, 0, sizeof e );
		scanf( "%d", &n );
		for( i=0; i<n; i++ )
		{
			scanf( "%d:%d %d %d %d %d", &a, &b, x1+i, y1+i, x2+i, y2+i );
			begin[i] = a*60 + b;
			end[i] = begin[i] + dis( x1[i], y1[i], x2[i], y2[i] );
		}
		for( i=0; i<n; i++ )
		for( j=i+1; j<n; j++ )
			if( end[i] + dis( x2[i], y2[i], x1[j], y1[j] ) < begin[j] )
				e[i][j] = true;
		printf( "%d\n", n - maxmatch( n, n, e, 0 ) );
	}
	return 0;
}