bzoj1620 [Usaco2008 Nov]Time Management 时间管理

最新推荐文章于 2024-11-20 09:24:36 发布

zhb1997

最新推荐文章于 2024-11-20 09:24:36 发布

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本文介绍了一个简单的工作调度问题：如何安排不同工作以确保按时完成。通过将任务按截止时间降序排列，并计算最大可行开始时间，文章提供了一种有效的方法来确定最晚可以何时开始工作以满足所有截止期限。

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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.

水题一道……

首先按deadline降序排一遍，保存一个当前的最大开始时间，那么当加入一个工作，要么保存答案的比这个工作的deadline大，要么后者大。无论如何一定要保证做完这个工作之后的时间比两者都大，这样才满足条件。所以两者取小的就行了。不会的自己再yy吧……

#include<cstdio>
#include<algorithm>
using namespace std;
struct work{
	int s,t;
}a[1010];
int n,ans=10000000;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline int min(int a,int b)
{return a<b?a:b;}
inline bool cmp(const work &a,const work &b)
{return a.s>b.s;}
int main()
{
	n=read();
	for (int i=1;i<=n;i++)
	{
		a[i].t=read();
		a[i].s=read();
	}
	sort(a+1,a+n+1,cmp);
	for (int i=1;i<=n;i++)
	  ans=min(ans,a[i].s)-a[i].t;
	if (ans<0) ans=-1;
	printf("%d",ans);
}