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.
1/* ***********************************************
2Author :111qqz
3Created Time :2016年04月04日 星期一 01时36分31秒
4File Name :code/bzoj/1620.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=1E3+7;
34pi a[N];
35int n;
36int main()
37{
38 #ifndef ONLINE_JUDGE
39 freopen("code/in.txt","r",stdin);
40 #endif
41 cin>>n;
42 for ( int i = 1 ; i <= n ; i++)
43 {
44 int x,y;
45 cin>>x>>y;
46 a[i]=make_pair(y,x);
47 }
48
49 sort(a+1,a+n+1);
50 int sum = 0 ;
51 int ans = inf;
52 for ( int i = 1 ; i <= n ; i++)
53 {
54 sum += a[i].sec;
55 ans = min(ans,a[i].fst-sum);
56 }
57 if (ans<0) ans=-1;
58 cout<<ans<<endl;
59
60 #ifndef ONLINE_JUDGE
61 fclose(stdin);
62 #endif
63 return 0;
64}