NYOJ 16 matrix nested

Rectangular nested

time limit:3000MS | Memory Limit:65535 KB

Difficulty:4

description

There are n rectangles, each rectangle can be described by A, B, indicating long and wide. Rectangular X (A, B) can be nested in rectangular Y (C, D) and is only used in rectangular Y (C, D) and is only when A <C, B <D, or B <C, A <D (corresponding to rotating X90 degrees). For example, (1, 5) can be nested in (6, 2), but cannot be nested in (3, 4). Your task is to select as many rectangles as possible, so that in addition to the last one, each rectangle can be nested in the next rectangle.

input

The first line is a positive number N (0 <n <10), indicating the number of test data groups,
The first line of each set of test data is a positive positive number n, indicating the number of rectangles in the group test data (n <= 1000)
Subsequent N rows, there are two numbers A, B (0 <a, b <100) per line, indicating the length and wide of the rectangle

output

Each set of test data outputs a number, indicating the number of rectangles that meet the conditions, and each group is output.

Sample input

sample output

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
struct space
{
	int a;
	int b;
}c[1005];
BOOL CMP (Space _x, space _y) / / must be sorted 
{
	if(_x.a==_y.a)
	return _x.b<_y.b;
	else 
	return _x.a<_y.a;
}
int main()
{
	int N,n,p,q,i,j,dp[1005];
	scanf("%d",&N);
	while(N--)
	{
		memset(dp,0,sizeof(dp));
		scanf("%d",&n);
		for(i=0;i<n;i++)
		{
			scanf("%d%d",&p,&q);
			c[i].a=p>q?p:q;
			c[i].b=p<q?p:q;
		}
		sort(c,c+n,cmp);
		int k=1;
		for(i=0;i<n;i++)
		{
			dp[i]=1;
			for(j=0;j<i;j++)
			{
				if(c[j].b<c[i].b&&c[j].a<c[i].a&&dp[j]+1>dp[i])
				dp[i]=dp[j]+1;
			}
			if(k<dp[i])
			k=dp[i];
		}
		printf("%d\n",k);
	}
	
	return 0;
}

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