HDU 5240 Exam （好水的题）

Exam

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 703    Accepted Submission(s): 354

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 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

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

 

    

Sample Output

Case #1: NOCase #2: YES

 

   一上来就发现是一道水题，要参加好多考试，给出复习一门课需要的时间，距离考试还剩下的时间，考试进行的时间，算是否能复习完。直接看代码就能懂

  注意：需要减掉上一次考试所拥有的复习时间

   

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>using namespace std;struct node{    int r,e,l;};node p[100500];bool cmp(node a,node b){    return a.e<b.e;}int main(){    int n,m,i,j,k=1,sum;    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(i=0;i<n;i++)        {            scanf("%d%d%d",&p[i].r,&p[i].e,&p[i].l);        }        sort(p,p+n,cmp);        sum=p[0].e-p[0].r;        if(sum<0)        {            printf("Case #%d: NO\n",k++);            continue;        }        else        {            for(i=1;i<n;i++)            {                sum+=p[i].e;                sum-=p[i-1].l;                sum-=p[i].r;                sum-=p[i-1].e;                if(sum<0)                {                    printf("Case #%d: NO\n",k++);                    break;                }            }            if(sum>0)            {                printf("Case #%d: YES\n",k++);            }        }    }    return 0;}