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

2016-04-03 · 2 min read · greedy

1620: [Usaco2008 Nov]Time Management 时间管理

Time Limit: 5 Sec  Memory Limit: 64 MB Submit: 636  Solved: 387 [Submit][Status][Discuss]

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.

思路：贪心。一眼题。5分钟A掉，爽。 按照s[i]排序，然后完成前i个任务的最晚开始时间为 s[i]-sum ，sum为t[1]+...+t[i]..

如果ans<0,说明无法完成，赋值为-1.

/* ***********************************************
Author :111qqz
Created Time :2016年04月04日 星期一 01时36分31秒
File Name :code/bzoj/1620.cpp
************************************************ */

 1#include <cstdio>
 2#include <cstring>
 3#include <iostream>
 4#include <algorithm>
 5#include <vector>
 6#include <queue>
 7#include <set>
 8#include <map>
 9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair

 1using namespace std;
 2const double eps = 1E-8;
 3const int dx4[4]={1,0,0,-1};
 4const int dy4[4]={0,-1,1,0};
 5const int inf = 0x3f3f3f3f;
 6const int N=1E3+7;
 7pi a[N];
 8int n;
 9int main()
10{
11	#ifndef  ONLINE_JUDGE 
12	freopen("code/in.txt","r",stdin);
13  #endif
14	cin>>n;
15	for ( int i = 1 ; i <= n ; i++)
16	{
17	    int x,y;
18	    cin>>x>>y;
19	    a[i]=make_pair(y,x);
20	}

 1	sort(a+1,a+n+1);
 2	int sum = 0 ;
 3	int ans = inf;
 4	for ( int i = 1 ; i <= n ; i++)
 5	{
 6	    sum += a[i].sec;
 7	    ans = min(ans,a[i].fst-sum);
 8	}
 9	if (ans<0) ans=-1;
10	cout<<ans<<endl;

1  #ifndef ONLINE_JUDGE  
2  fclose(stdin);
3  #endif
4    return 0;
5}