Ever the maturing businessman, Farmer John realizes that he must manage his time effectively. He has N jobs conveniently numbered 1..N (1 <= N <= 1,000) to accomplish (like milking the cows, cleaning the barn, mending the fences, and so on). To manage his time effectively, he has created a list of the jobs that must be finished. Job i requires a certain amount of time T_i (1 <= T_i <= 1,000) to complete and furthermore must be finished by time S_i (1 <= S_i <= 1,000,000). Farmer John starts his day at time t=0 and can only work on one job at a time until it is finished. Even a maturing businessman likes to sleep late; help Farmer John determine the latest he can start working and still finish all the jobs on time.
N个工作,每个工作其所需时间,及完成的Deadline,问要完成所有工作,最迟要什么时候开始.
Line 1: A single integer: N
Lines 2..N+1: Line i+1 contains two space-separated integers: T_i and S_i
3 5
8 14
5 20
1 16
Farmer John has 4 jobs to do, which take 3, 8, 5, and 1 units of
time, respectively, and must be completed by time 5, 14, 20, and
16, respectively.
2
Farmer John must start the first job at time 2. Then he can do
the second, fourth, and third jobs in that order to finish on time.
#include
#include
#include
using namespace std;
int n;
int start;
struct node{
int tm,ed;
};
node edge[1005];
bool cmp(node a,node b)
{
return a.ed>b.ed;
}
int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d%d",&edge[i].tm,&edge[i].ed);
start=max(start,edge[i].ed);
}
sort(edge+1,edge+n+1,cmp);
for(int i=1;i<=n;i++)
{
if(start>edge[i].ed)
start=edge[i].ed;
start-=edge[i].tm;
if(start<0)
{
printf("-1");
exit(0);
}
}
printf("%d",start);
}