# BZOJ 1620: [Usaco2008 Nov]Time Management 时间管理

#### Description

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.

#### Input

* Line 1: A single integer: N

* Lines 2..N+1: Line i+1 contains two space-separated integers: T_i and S_i

#### Output

* Line 1: The latest time Farmer John can start working or -1 if Farmer John cannot finish all the jobs on time.

end.

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
//by zrt
//problem:
using namespace std;
int n;
struct N{
int t,s;
friend bool operator < (N a,N b){
return a.s<b.s;
}
}a[1005];
int minn=1<<30;
bool judge(int start){
int t=start;
for(int i=1;i<=n;i++){
if(t+a[i].t>a[i].s) return 0;
else t+=a[i].t;
}
return 1;
}
int main(){
#ifdef LOCAL
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
scanf("%d",&n);
for(int i=1;i<=n;i++){
scanf("%d%d",&a[i].t,&a[i].s);
minn=min(minn,a[i].s);
}
sort(a+1,a+n+1);
int l=0,r=minn;
if(!judge(0)){
puts("-1");
goto ed;
}
while(r-l>1){
int m=(l+r)>>1;
if(judge(m)){
l=m;
}else r=m;
}
printf("%d\n",l);
ed:;return 0;
}
posted @ 2014-09-10 13:41  zrt  阅读(167)  评论(0编辑  收藏  举报

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