HDOJ 5240 Exam

Exam

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1112    Accepted Submission(s): 550

Problem 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 leastri 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 integern≤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 is the number of test cases, and ans is ''YES'' (without quotes) if DRD can pass all the courses, and otherwise ''NO'' (without quotes).x

 

 

 
Sample Input
233 2 25 100 27 1000 2
33 10 25 100 27 1000 2

 

 

Sample Output

Case #1: NOCase #2: YES

 

 

Source

The 2015 ACM-ICPC China Shanghai Metropolitan Programming Contest

 

 一开始没有想过结束时间会不排序的问题，之后有人讲了才发现这点~然后又因为数型wa了两次

醉醉的。

 

#include<stdio.h>#include<algorithm>using namespace std;struct re{long long s,e,l;}a[100001];int cmp(re a,re b){return a.e<b.e;}int main(){int t,i,n;int k=1;scanf("%d",&t);while(t--){int flag=1;scanf("%d",&n);for(i=0;i<n;i++)scanf("%lld%lld%lld",&a[i].s,&a[i].e,&a[i].l);sort(a,a+n,cmp);for(i=0;i<n;i++){if(a[i].s>a[i].e){flag=0;break;}a[i+1].s+=a[i].l+a[i].s;}if(flag)printf("Case #%d: YES\n",k++);elseprintf("Case #%d: NO\n",k++);}return 0;}