P2920 [USACO08NOV]Time Management S

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.

作为一名忙碌的商人，约翰知道必须高效地安排他的时间.他有N工作要 做，比如给奶牛挤奶，清洗牛棚，修理栅栏之类的.

为了高效，列出了所有工作的清单.第i分工作需要T_i单位的时间来完成，而 且必须在S_i或之前完成.现在是0时刻.约翰做一份工作必须直到做完才能停 止.

所有的商人都喜欢睡懒觉.请帮约翰计算他最迟什么时候开始工作，可以让所有工作按时完成.（如果无法完成全部任务，输出-1）

输入格式

* Line 1: A single integer: N

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

输出格式

* 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

输入 #1复制

4 
3 5 
8 14 
5 20 
1 16 

输出 #1复制

2 

说明/提示

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.

【题目分析】 这个题就是一个二分+模拟的题目，首先我们终止时间早的肯定要先做，这样的话，就需要按照区间终点进行排序，接下来我们枚举从哪个点开始可以完全把所有的工作都做完，如果当前时间可以，我们就将时间往后挪动，否则就往前移动，直到找到答案位置。

【代码实现】

#include
using namespace std;
struct Node
{
	int l,r;
	bool operator<(const Node& w)const
	{
		return r>n;
	for(int i=1;i<=n;i++)
	{
		cin>>node[i].l>>node[i].r;
	}
	sort(node+1,node+n+1);
	int l=-1,r=node[1].r;//答案区间 
	while(l>1;
		if(check(mid)) l=mid;
		else r=mid-1;
	}
	cout<