zoukankan      html  css  js  c++  java
  • POJ-2376

    Cleaning Shifts
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 23483   Accepted: 5872

    Description

    Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided the day into T shifts (1 <= T <= 1,000,000), the first being shift 1 and the last being shift T. 

    Each cow is only available at some interval of times during the day for work on cleaning. Any cow that is selected for cleaning duty will work for the entirety of her interval. 

    Your job is to help Farmer John assign some cows to shifts so that (i) every shift has at least one cow assigned to it, and (ii) as few cows as possible are involved in cleaning. If it is not possible to assign a cow to each shift, print -1.

    Input

    * Line 1: Two space-separated integers: N and T 

    * Lines 2..N+1: Each line contains the start and end times of the interval during which a cow can work. A cow starts work at the start time and finishes after the end time.

    Output

    * Line 1: The minimum number of cows Farmer John needs to hire or -1 if it is not possible to assign a cow to each shift.

    Sample Input

    3 10
    1 7
    3 6
    6 10

    Sample Output

    2

    Hint

    This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed. 

    INPUT DETAILS: 

    There are 3 cows and 10 shifts. Cow #1 can work shifts 1..7, cow #2 can work shifts 3..6, and cow #3 can work shifts 6..10. 

    OUTPUT DETAILS: 

    By selecting cows #1 and #3, all shifts are covered. There is no way to cover all the shifts using fewer than 2 cows.

    题意:

    给出n个区间,求覆盖1~n的最小区间数。

    将区间按左区间最小排序,若左区间相等,则将右区间最大优先。

    选取区间最大的。

     1 //#include<bits/stdc++.h>
     2 #include<cstdio>  
     3 #include<cstring>  
     4 #include<algorithm>
     5 #include<iostream>
     6 using namespace std;
     7 
     8 struct node{
     9     int x;
    10     int y;
    11 }a[30000];
    12 
    13 int cmp(node a,node b){
    14     if(a.x!=b.x) 
    15     return a.x<b.x;
    16     else
    17     return a.y>b.y;
    18 }
    19 
    20 int main(){
    21     ios::sync_with_stdio(false);
    22     int n,t;
    23     while(cin>>n>>t&&n&&t){
    24         for(int i=0;i<n;i++){
    25             cin>>a[i].x>>a[i].y;
    26         }
    27         sort(a,a+n,cmp);
    28         if(a[0].x!=1){
    29             cout<<"-1"<<endl;
    30             continue;
    31         }
    32         int temp=0,k=1,ri=a[0].y;
    33         for(int i=1;i<n;i++){
    34             if(a[i].x>ri+1){//前一个即为可选区间 
    35                 k++;
    36                 ri=temp;
    37             }
    38             if(a[i].x<=ri+1){
    39                 if(a[i].y>temp){
    40                     temp=a[i].y;//更新最右区间 
    41                 }
    42                 if(a[i].y==t){
    43                     k++;
    44                     ri=t;
    45                     break;
    46                 }
    47             }
    48         }
    49         if(ri==t){
    50             cout<<k<<endl;
    51         }
    52         else{
    53             cout<<"-1"<<endl;
    54         }
    55     }
    56     return 0;
    57 }
  • 相关阅读:
    PyCharm黄色波浪线提示: Simplify chained comparison
    SQL Server 2017 新功能分享
    阿里云RDS for SQL Server使用的一些最佳实践
    Spark入门PPT分享
    亿级SQL Server运维的最佳实践PPT分享
    使用T-SQL找出执行时间过长的作业
    SQL Server中TOP子句可能导致的问题以及解决办法
    广州的小伙伴福利-由微软组织的在广州SQL Server线下活动
    在SQL Server中为什么不建议使用Not In子查询
    微软Ignite大会我的Session(SQL Server 2014 升级面面谈)PPT分享
  • 原文地址:https://www.cnblogs.com/Kiven5197/p/7298243.html
Copyright © 2011-2022 走看看