zoukankan      html  css  js  c++  java
  • 8.6贪心策略例题:区间覆盖问题

    /**
    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

    */
    //POJ2376

    找每一个子问题的最优解

     1 import java.util.Arrays;
     2 import java.util.Scanner;
     3 
     4 public class Eight_6贪心策略例题_区间覆盖问题 {
     5     private static class Job implements Comparable<Job> {
     6         int s;
     7         int t;
     8 
     9         public Job(int s, int t) {
    10             super();
    11             this.s = s;
    12             this.t = t;
    13         }
    14 
    15         @Override
    16         public int compareTo(Job Other) {
    17             int x = this.s-Other.s;
    18             if(x == 0)
    19                 return this.t-Other.t;
    20             else
    21                 return x;
    22         }
    23 
    24         @Override
    25         public String toString() {
    26             return "Job [s=" + s + ", t=" + t + "]";
    27         }    
    28     }
    29     
    30     public static void main(String[] args) {
    31         Scanner in = new Scanner(System.in);
    32         int N = in.nextInt();
    33         int T = in.nextInt();
    34         Job[] jobs = new Job[N];
    35         for(int i = 0; i < N; i++){
    36             jobs[i] = new Job(in.nextInt(), in.nextInt());
    37         }
    38         
    39         Arrays.sort(jobs);
    40         int start = 1; //要覆盖的目标点,end覆盖该点的所有区间中右端点最右
    41         int end = 1;
    42         int ans = 1;
    43         for(int i = 0; i < T; i++){
    44             int s = jobs[i].s;
    45             int t = jobs[i].t;
    46             
    47             if(i == 0 && s > 1)
    48                 break;
    49             if(s <= start){
    50                 end = Math.max(t, end);
    51             }else{ //开始下一个区间
    52                 ans++;
    53                 start = end+1; //更新起点
    54                 if(s <= start){
    55                     end = Math.max(t, end);
    56                 }else{
    57                     break;
    58                 }
    59             }
    60             if(end >= T) //区间右端点大于输入段退出循环
    61                 break;
    62         }
    63         
    64         if(end < T)
    65             System.out.println("-1");
    66         else
    67             System.out.println(ans);
    68     }
    69 }
  • 相关阅读:
    51nod 1494 选举拉票 | 线段树
    51nod 1295 XOR key | 可持久化Trie树
    Codeforces 438D (今日gg模拟第二题) | 线段树 考察时间复杂度的计算 -_-|||
    51nod 1563 坐标轴上的最大团(今日gg模拟第一题) | 线段覆盖 贪心 思维题
    良心的可持久化线段树教程
    51nod 1593 公园晨跑 | ST表(线段树?)思维题
    51nod 1595 回文度 | 马拉车Manacher DP
    51nod 1522 上下序列
    胡小兔的OI日志3 完结版
    51nod 1510 最小化序列 | DP 贪心
  • 原文地址:https://www.cnblogs.com/z1110/p/12761926.html
Copyright © 2011-2022 走看看