Rectangles dp

最新推荐文章于 2020-11-02 18:47:05 发布

Rachel-Zhang

最新推荐文章于 2020-11-02 18:47:05 发布

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文章标签： integer struct input each pair output

本文链接：https://blog.youkuaiyun.com/abcjennifer/article/details/5742485

Rectangles

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 194 Accepted Submission(s): 76

Problem Description

A rectangle in the Cartesian plane is speci ed by a pair of coordinates (x1 , y1) and (x2 , y2) indicating its lower-left and upper-right corners, respectively (where x1 ≤ x2 and y1 ≤ y2). Given a pair of rectangles,A = ((x^A₁ , y^A₁ ), (x^A₂ ,y^A₂ )) and B = ((x^B₁ , y^B₁ ), (x^B₂ , y^B₂ )), we write A ≤ B (i.e., A "precedes" B), if x^A₂ < x^B₁ and y^A₂ < y^B₁ :In this problem, you are given a collection of rectangles located in the two-dimension Euclidean plane. Find the length L of the longest sequence of rectangles (A₁,A₂,…,A_L) from this collection such that A₁ ≤ A₂ ≤ … ≤ A_L.

Input

The input fi le will contain multiple test cases. Each test case will begin with a line containing a single integer n (where 1 ≤ n ≤ 1000), indicating the number of input rectangles. The next n lines each contain four integers xⁱ₁ ,yⁱ₁ ,xⁱ₂ ,yⁱ₂ (where -1000000 ≤ xⁱ₁ ≤ xⁱ₂ ≤ 1000000, -1000000 ≤ yⁱ₁ ≤ yⁱ₂ ≤ 1000000, and 1 ≤ i ≤ n), indicating the lower left and upper right corners of a rectangle. The end-of-file is denoted by asingle line containing the integer 0.

Output

For each input test case, print a single integer indicating the length of the longest chain.

Sample Input

3
1 5 2 8
3 -1 5 4
10 10 20 20
2
2 1 4 5
6 5 8 10
0

Sample Output

2
1
#include<iostream>
using namespace std;
int n;
#define M 1010
#define INF 100000000
int len[M];
int max;
int longest=0;
struct RECTANGLE
{
	int x1,y1,x2,y2;
}rec[M];

int comp(const void *a,const void *b)
{
	struct RECTANGLE *aa=(struct RECTANGLE*)a;
	struct RECTANGLE *bb=(struct RECTANGLE*)b;
	if(aa->x1!=bb->x1)
		return aa->x2-bb->x2;
	return aa->y2-bb->y2;
}

void dp(int n)
{
	int i,j,max;
	for(i=0;i<n;i++) 
	{
		max = 1;
		len[i] = 1;
		for(j=0;j<i;j++)
		{
			if(rec[j].x2<rec[i].x1&&rec[j].y2<rec[i].y1)
				max=len[j]+1;
			if(len[i]<max)
				len[i]=max;
		}
		if(len[i]>longest)
			longest=len[i];
	}
	printf("%d/n", longest);
}

int main()
{
	while(cin>>n&&n)
	{
		longest=0;
		for(int i=0;i<n;i++)
			cin>>rec[i].x1>>rec[i].y1>>rec[i].x2>>rec[i].y2;
		qsort(rec,n,sizeof(RECTANGLE),comp);
		dp(n);
	}
}