HDOJ 5240 Exam(水)

Exam

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1031    Accepted Submission(s): 513


[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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.

 

[align=left]Output[/align]
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).

 

[align=left]Sample Input[/align]

2
3
3 2 2
5 100 2
7 1000 2
3
3 10 2
5 100 2
7 1000 2

 

[align=left]Sample Output[/align]

Case #1: NO
Case #2: YES

 

 

题意：有n门课要考试了，接下来n行每行有三个数字r,e,l。r表示当前这门课需要复习的时间，e表示从现在开始距离 当前这门课考试还有多长时间，l表示当前这门课的上课时间。问着n门课是否都能复习并完成考试。

很水的题，不过要注意给这些科目按照考试先后排序，数据不一定是升序。这个地方wa了三发，实在可惜。

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
int t,e,l;
}num[100010];

int cmp(node a,node b)
{
return a.e<b.e;
}

int main()
{
int T,i,e,n,k=1;
__int64 t,l;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(i=0;i<n;++i)
scanf("%d%d%d",&num[i].t,&num[i].e,&num[i].l);
sort(num,num+n,cmp);
int sign=1,time=0;
for(i=0;i<n;++i)
{
time+=num[i].t;
if(time>num[i].e)
{
sign=0;
break;
}
time+=num[i].l;
}
if(sign)
printf("Case #%d: YES\n",k++);
else
printf("Case #%d: NO\n",k++);
}
return 0;
}