hdoj 5240 Exam【贪心】

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Exam

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1067 Accepted Submission(s): 529

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

Source

The 2015 ACM-ICPC China Shanghai Metropolitan Programming Contest

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;struct node{ll re,be,last;//re复习时间,be考试开始时间,last考试持续时间; };//考试开始时间一定要在复习时间之前！ node p[110000];bool cmp(node a, node b){return a.be < b.be;//按考试开始时间先后排序 }int main(){int t,ca=1;scanf("%d", &t);while(t--){int n;scanf("%d", &n);//n门考试 int i,j,k;for(i = 0; i < n; i++){scanf("%lld%lld%lld", &p[i].re, &p[i].be, &p[i].last);}sort(p,p+n,cmp);//排序 int flag = 1;int a = p[0].be;//a 两场考试之间间隔时间 for(i = 0; i < n; i++){if( i &&p[i].be < p[i-1].be+p[i-1].last){flag = 0; break;//考试时间冲突 无法通过所有考试 }if(i){a+=(p[i].be - p[i-1].be-p[i-1].last);//时间间隔 }if(a < p[i].re){//复习时间在考试开始之后，就无法通过这个考试，直接退出 flag = 0;break;}else{a-=p[i].re;//考试之间的时间间隔减去复习时间,可以接着复习下一门. }}printf("Case #%d: ",ca++);if(flag)printf("YES\n");elseprintf("NO\n");}return 0;}

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