1620: [Usaco2008 Nov]Time Management 时间管理

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

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.

 

开始想的是线段树，但是觉得太麻烦，后来觉得二分也可以..也觉得麻烦...然后是贪心，只要从deadline从大到小添加就好了，如果当前deadline之前要进行别的任务，那就从那个任务的开始点往前....

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define MAX 100010
#define ll long long
#define inf 600000000
using namespace std;
struct data{
    int d,w;
}a[MAX];
bool cmp(data a,data b){
    return a.d>b.d;
}
int main(){
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%d%d",&a[i].w,&a[i].d);
    }
    sort(a+1,a+n+1,cmp); 
    int ans=inf; 
    for (int i=1;i<=n;i++)  
      ans=min(ans,a[i].d)-a[i].w;  
    if (ans<0) ans=-1;  
    printf("%d",ans);  
}