【BZOJ】1620: [Usaco2008 Nov]Time Management 时间管理（贪心）

最新推荐文章于 2025-08-20 11:37:09 发布

转载最新推荐文章于 2025-08-20 11:37:09 发布 · 64 阅读

文章标签：

#php

本文介绍了一个简单而有效的任务调度算法，用于解决多个有截止时间的任务如何安排的问题。通过将任务按截止时间逆序排列并反向计算最早开始时间，可以得出完成所有任务所需的最晚开始时间。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

http://www.lydsy.com/JudgeOnline/problem.php?id=1620

一开始想不通啊。。

其实很简单。。。

每个时间都有个完成时间，那么我们就从最大的完成时间的开始往前推

每一次更新最早开始时间（min(ans, a[i].y)代表i事件最早的完成时间）

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << #x << " = " << x << endl
#define printarr(a, n, m) rep(aaa, n) { rep(bbb, m) cout << a[aaa][bbb]; cout << endl; }
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
inline const int max(const int &a, const int &b) { return a>b?a:b; }
inline const int min(const int &a, const int &b) { return a<b?a:b; }

const int N=1005;
struct data { int x, y; }a[N];
inline bool cmp(const data &x, const data &y) { return x.y>y.y; }
int main() {
	int n=getint();
	for1(i, 1, n) read(a[i].x), read(a[i].y);
	sort(a+1, a+1+n, cmp);
	int ans=~0u>>1;
	for1(i, 1, n) ans=min(ans, a[i].y)-a[i].x;
	if(ans<0) puts("-1");
	else print(ans);
	return 0;
}

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.