1620: [Usaco2008 Nov]Time Management 时间管理
Time Limit: 5 Sec Memory Limit: 64 MBSubmit: 571 Solved: 343[][][]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
Sample Output
HINT
Source
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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