NC210102. 时间管理
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.
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.
输入描述
* Line 1: A single integer: N
* Lines 2..N+1: Line i+1 contains two space-separated integers: T_i and S_i
输出描述
* Line 1: The latest time Farmer John can start working or -1 if Farmer John cannot finish all the jobs on time.
示例1
输入:
4 3 5 8 14 5 20 1 16
输出:
2
C++(clang++ 11.0.1) 解法, 执行用时: 3ms, 内存消耗: 388K, 提交时间: 2022-08-10 20:26:21
#include<bits/stdc++.h>
using namespace std;
struct sj
{
int s;
int j;
}a[1010];
bool cmp(sj a,sj b)
{
return a.j<b.j;
}
int main()
{
int n,l=0,r,mid,i,c,f,da=-1;
cin>>n;
for(i=1;i<=n;i++)
cin>>a[i].s>>a[i].j;
sort(a+1,a+n+1,cmp);
r=a[n].j;
while(l<=r)
{
mid=l+(r-l)/2,c=mid,f=1;
for(i=1;i<=n;i++)
{
if(c+a[i].s>a[i].j)
{
f=0;
break;
}
c+=a[i].s;
}
if(f==1)
{
da=mid;
l=mid+1;
}
else
r=mid-1;
}
cout<<da;
return 0;
}