HDOJ 5240 Exam 【简单模拟】

Exam

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1041    Accepted Submission(s): 521

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

恩，题目大意就是说，预习一科目所需时间，多久后考试开始，考试持续时间，问能否在所有时间不冲突的情况下完成所有科目，就排序然后更新一下（个人看法）

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#define maxn 100000+10using namespace std;struct Node{    int t,s,l;};Node node[maxn];int cmp(Node a,Node b){    return a.s<b.s;}int main(){    int t,n,cnt=0;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        for(int i=0;i<n;++i)            scanf("%d%d%d",&node[i].t,&node[i].s,&node[i].l);        sort(node,node+n,cmp);        int flag=1;        for(int i=0;i<n;++i)        {            if(node[i].t>node[i].s)            {                flag=0;                break;            }            node[i+1].t+=node[i].t+node[i].l;        }        printf("Case #%d: ",++cnt);        if(flag)            printf("YES\n");        else            printf("NO\n");    }    return 0;}