hdoj 5240 Exam

Exam

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1034    Accepted Submission(s): 514

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), wherex 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

233 2 25 100 27 1000 233 10 25 100 27 1000 2

 

Sample Output

Case #1: NOCase #2: YES

 

题意：一个人有  n  门课要考，每一门课给出复习所必须花费的时间  r  ，这门课考试距现在所剩的时间  e  ，考这门课时所用的时间  l  。问这个人是否能通过全部的考试。

思路：现对考试距现在所剩的时间  e  从小到大进行排序，如果  e  相等，则将  l  进行从小到大排序。然后对复习时间与考试时间进行模拟。

代码：

#include<stdio.h>#include<algorithm>using namespace std;struct record{int r,e,l;}num[100010];bool cmp(record a,record b){return a.e<b.e;}int main(){int T,n;int i,k=1;scanf("%d",&T);while(T--){scanf("%d",&n);int f=0;for(i=0;i<n;i++){scanf("%d%d%d",&num[i].r,&num[i].e,&num[i].l);     if(num[i].r>num[i].e)    {    f=1;    }}if(f==1){printf("Case #%d: NO\n",k++);continue;}sort(num,num+n,cmp);int sum=0;for(i=1;i<n;i++){sum+=num[i-1].r+num[i-1].l;if(num[i].e<num[i].r+sum){f=1;break;}}if(f==1)printf("Case #%d: NO\n",k++);elseprintf("Case #%d: YES\n",k++);}return 0;}