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.
题解:
开始时间有单调性,如果从时间t开始能完成,那么从时间t’<t都能完成。
所以我们可以二分一个t。
在t确定后根据贪心每次做Deadline最靠前的任务，模拟一下。
注意无解输出-1。囧
end.
代码：
#include<cstdio> #include<cstring> #include<algorithm> #include<queue> //by zrt //problem: using namespace std; int n; struct N{ int t,s; friend bool operator < (N a,N b){ return a.s<b.s; } }a[1005]; int minn=1<<30; bool judge(int start){ int t=start; for(int i=1;i<=n;i++){ if(t+a[i].t>a[i].s) return 0; else t+=a[i].t; } return 1; } int main(){ #ifdef LOCAL freopen("in.txt","r",stdin); freopen("out.txt","w",stdout); #endif scanf("%d",&n); for(int i=1;i<=n;i++){ scanf("%d%d",&a[i].t,&a[i].s); minn=min(minn,a[i].s); } sort(a+1,a+n+1); int l=0,r=minn; if(!judge(0)){ puts("-1"); goto ed; } while(r-l>1){ int m=(l+r)>>1; if(judge(m)){ l=m; }else r=m; } printf("%d ",l); ed:;return 0; }