# bzoj 1620 管理时间

## 题目描述

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.

## 输入输出格式

* Line 1: A single integer: N

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

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

## 输入输出样例

4
3 5
8 14
5 20
1 16


2


## 说明

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.

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.

 1 #include<bits/stdc++.h>
2 using namespace std;
3 #define re register
4 #define R  re int
5 #define rep(i,a,b) for(R i=a;i<=b;i++)
6 #define Rep(i,a,b) for(R i=a;i>=b;i--)
7 #define ms(i,a)    memset(a,i,sizeof(a))
9   x=0; char c=0;
10   while (!isdigit(c)) c=getchar();
11   while (isdigit(c)) x=x*10+(c^48),c=getchar();
12 }
13 int const N=1001;
14 struct node{
15   int s,t;
16   bool operator < ( const node &rhs) const{
17     return s<rhs.s;
18   }
19 }a[N];
20 int n;
21 int main(){
24   sort(a+1,a+n+1);
25   int ans=2000000;
26   Rep(i,n,1){
27     if(a[i].s>ans) ans=ans-a[i].t;
28     else ans=a[i].s-a[i].t;
29   }
30   if(ans<0) ans=-1;
31   printf("%d\n",ans);
32   return 0;
33 }
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posted @ 2018-12-29 17:04  zjxxcn  阅读(130)  评论(0编辑  收藏  举报