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.
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains two space-separated integers: T_i and S_i
* Line 1: The latest time Farmer John can start working or -1 if Farmer John cannot finish all the jobs on time.
Silver
先排序一下，注意是排序完成的时间，然后用二分答案就ok了，感觉就是好像如果能用枚举的话用二分答案搞。
我草草草草一直wa居然是因为一个点需要输出-1！！！看题看题看题看题！！！
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#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int n;
struct edge{
int dis,to;
bool operator<(const edge&rhs) const{
return to<rhs.to;}
};
edge edges[1005];
bool pd(int x){
int ans=x;
for(int i=1;i<=n;i++){
edge &s=edges[i];
ans+=s.dis;
if(ans>s.to)
return false;
}
return true;
}
int main(){
scanf("%d",&n);
int ans=-2;
int l=0;int r=0x7fffffff;
for(int i=1;i<=n;i++){
scanf("%d%d",&edges[i].dis,&edges[i].to);
r=min(r,edges[i].to);
}
sort(edges+1,edges+n+1);
while(l<=r){
int m=(l+r)/2;
if(pd(m)){
ans=m,l=m+1;
}
else
r=m-1;
}
if(ans!=-2)
printf("%d\n",ans);
else
printf("-1");
return 0;
}
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