# [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.

## 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

4
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

2

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;
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(){
for(int i = 1; i <= N; i++){
}
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;
}

posted @ 2017-09-11 19:52  Elder_Giang  阅读(103)  评论(0编辑  收藏  举报