Hdu 5240 Exam【贪心】

Description

As this term is going to end, DRD needs to prepare for his final exams. 

DRD has n exams. They are all hard, but their difficulties are different. DRD will spend at least ri hours on the i-th course before its exam starts, or he will fail it. The i-th course's exam will take place ei hours later from now, and it will last for li hours. When DRD takes an exam, he must devote himself to this exam and cannot (p)review any courses. Note that DRD can review for discontinuous time. 

So he wonder whether he can pass all of his courses. 

No two exams will collide. 

 

Input

First line: an positive integer T≤20 indicating the number of test cases. 
There are T cases following. In each case, the first line contains an positive integer n≤105, and n lines follow. In each of these lines, there are 3 integers ri,ei,li, where 0≤ri,ei,li≤109. 

 

Output

For each test case: output ''Case #x: ans'' (without quotes), where x is the number of test cases, and ans is ''YES'' (without quotes) if DRD can pass all the courses, and otherwise ''NO'' (without quotes). 

 

Sample Input

2
3
3 2 2
5 100 2
7 1000 2
3
3 10 2
5 100 2
7 1000 2 

 

Sample Output

Case #1: NO
Case #2: YES 

 

给出若干门课的通过需要复习的时间，考试开始时间和考试持续时间，求这个人能否参加和通过所有考试，贪心排序，看是否有重合的区间.......

水掉..

#include<stdio.h>
#include<algorithm>
using namespace std;
int n;
struct work
{
	int r,e,l;
}x[100005];

int cmp(work a,work b)
{
	if(a.e==b.e)
	{
		return a.r<b.r;
	}
	return a.e<b.e;
}

int judge()
{
	int time=0;
	for(int i=0;i<n;++i)
	{
		if(x[i].r+time<=x[i].e)//可以通过
		{
			time=x[i].e+x[i].l;//考完试了
		}
		else
		{
			return 0;
		}
	}
	return 1;
}

int main()
{
	int t;
	//freopen("shuju.txt","r",stdin);
	scanf("%d",&t);
	for(int k=1;k<=t;++k)
	{
		scanf("%d",&n);
		for(int i=0;i<n;++i)
		{
			scanf("%d%d%d",&x[i].r,&x[i].e,&x[i].l);
		}
		sort(x,x+n,cmp);
		if(judge())
		{
			printf("Case #%d: YES\n",k);
		}
		else
		{
			printf("Case #%d: NO\n",k);
		}
	}
	return 0;
}