# [BZOJ1620][Usaco2008 Nov]Time Management 时间管理

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

Time Limit: 5 Sec Memory Limit: 64 MB Submit: 842 Solved: 533 [Submit][Status][Discuss]## 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.

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.

## 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.

## Sample Input

3 5

8 14

5 20

1 16

INPUT DETAILS:

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.

## Sample Output

OUTPUT DETAILS:

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 <cstdio> #include <cstring> #include <algorithm> using namespace std; char buf[5000000], *ptr = buf - 1; inline int readint(){ int n = 0; char ch = *++ptr; while(ch < '0' || ch > '9') ch = *++ptr; while(ch <= '9' && ch >= '0'){ n = (n << 1) + (n << 3) + ch - '0'; ch = *++ptr; } return n; } const int maxn = 1000 + 10; struct Node{ int t, s; Node(){} bool operator < (const Node &x) const { return s < x.s; } }a[maxn]; int main(){ fread(buf, sizeof(char), sizeof(buf), stdin); int N = readint(); for(int i = 1; i <= N; i++){ a[i].t = readint(); a[i].s = readint(); } sort(a + 1, a + N + 1); int T = 1 << 30; for(int i = N; i; i--) T = min(T - a[i].t, a[i].s - a[i].t); if(T < 0) puts("-1"); else printf("%d\n", T); return 0; }