BZOJ 1620: [Usaco2008 Nov]Time Management 時間管理

1620: [Usaco2008 Nov]Time Management 時間管理

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 823  Solved: 520
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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個工作,每個工作其所需時間,及完成的Deadline,問要完成所有工作,最遲要什麼時候開始.

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

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.

HINT

Source

Silver

題解：按結束時間排序，從後往前掃一遍。。。然後輸出就好了。。。

代碼：

#include<bits/stdc++.h>
using namespace std;
int n,now=INT_MAX;
struct node{int x,y;}a[1001];
bool cmp(node a,node b){return a.y<b.y;}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	   scanf("%d%d",&a[i].x,&a[i].y);
	sort(a+1,a+n+1,cmp);
	for(int i=n;i>=1;i--)
		now=min(now,a[i].y)-a[i].x;
	if(now<0)printf("-1");
	else printf("%d",now);
	return 0;
}