Exam
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1028 Accepted Submission(s): 510
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.
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.
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
2
3
3 2 2
5 100 2
7 1000 2
3
3 10 2
5 100 2
7 1000 2
Sample Output
Case #1: NO
Case #2: YES
题解：简单贪心，模拟一下；
代码：
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cmath> 5 #include<algorithm> 6 #define mem(x,y) memset(x,y,sizeof(x)) 7 using namespace std; 8 const int INF=0x3f3f3f3f; 9 const int MAXN=1e5+100; 10 struct Node{ 11 int r,e,l; 12 }; 13 Node dt[MAXN]; 14 int cmp(Node a,Node b){ 15 return a.e<b.e; 16 } 17 int main(){ 18 int T,N,cnt=0; 19 scanf("%d",&T); 20 while(T--){ 21 scanf("%d",&N); 22 for(int i=0;i<N;i++) 23 scanf("%d%d%d",&dt[i].r,&dt[i].e,&dt[i].l); 24 sort(dt,dt+N,cmp); 25 int flot=1,tim=0; 26 for(int i=0;i<N;i++){ 27 tim+=dt[i].r; 28 if(tim>dt[i].e){ 29 flot=0;break; 30 } 31 tim+=dt[i].l; 32 } 33 if(!flot)printf("Case #%d: NO ",++cnt); 34 else printf("Case #%d: YES ",++cnt); 35 } 36 return 0; 37 }