zoukankan      html  css  js  c++  java
  • [USACO08NOV]时间管理Time Management

    题目描述

    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.

    作为一名忙碌的商人,约翰知道必须高效地安排他的时间.他有N工作要 做,比如给奶牛挤奶,清洗牛棚,修理栅栏之类的.

    为了高效,列出了所有工作的清单.第i分工作需要T_i单位的时间来完成,而 且必须在S_i或之前完成.现在是0时刻.约翰做一份工作必须直到做完才能停 止.

    所有的商人都喜欢睡懒觉.请帮约翰计算他最迟什么时候开始工作,可以让所有工作按时完 成.

    输入输出格式

    输入格式:
    • Line 1: A single integer: N

    • Lines 2..N+1: Line i+1 contains two space-separated integers: T_i and S_i
    输出格式:
    • Line 1: The latest time Farmer John can start working or -1 if Farmer John cannot finish all the jobs on time.

    输入输出样例

    输入样例#1: 复制
    4 
    3 5 
    8 14 
    5 20 
    1 16 
    
    输出样例#1: 复制
    2 
    

    说明

    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.

    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

    以时间限制s为关键字从大到小排序,贪心选取;

    代码

     1 #include<cstdio>
     2 #include<algorithm>
     3 using namespace std;
     4 const int maxn=1e3+10;
     5 inline int min_(int x,int y){return x<y?x:y;}
     6 int n,ans=1e6;
     7 struct nate{int t,s;}s[maxn];
     8 bool comp(nate x,nate y){return x.s>y.s;}
     9 int main(){
    10     scanf("%d",&n);
    11     for(int i=1;i<=n;i++) scanf("%d%d",&s[i].t,&s[i].s);
    12     sort(s+1,s+n+1,comp);
    13     for(int i=1,j=1e6;i<=n;i++) ans=min_(ans,s[i].s)-s[i].t;
    14     if(ans>=0) printf("%d
    ",ans);
    15     else puts("-1");
    16     return 0;
    17 }
  • 相关阅读:
    我认为的架构师
    Jenkins github账号密码验证方式失效 解决方式
    android逆向奇技淫巧二十一:ida反反调试&加密算法跟踪(未完待续)(六)
    android逆向奇技淫巧十九:unidbg模拟执行和trace x音so代码(四)
    android逆向奇技淫巧十八:x音so层代码花指令防护分析(三)
    android逆向奇技淫巧十七:android客户端自动x红包(一):代码原理分析
    测试工具安装汇总
    javascript事件节流和防抖
    CompletableFuture-更优雅的使用多线程
    青春
  • 原文地址:https://www.cnblogs.com/J-william/p/7725793.html
Copyright © 2011-2022 走看看