zoukankan      html  css  js  c++  java
  • ACM.hdu1025

    to get the ans of how many roads at most that can be built between two line without intersection of roads,we need sort the input sequence at ont edge and then deal with anther line with LIS

    way,meanwhile, not forget using the nlogn way.

     1 #include<iostream>
     2 #include<algorithm>
     3 using namespace std;
     4 const int N=500009;
     5 struct LINE
     6 {
     7     int from,to;    
     8     friend bool operator<(LINE A,LINE B)
     9     {
    10         return A.from<B.from;
    11     }
    12 };
    13 LINE node[N];
    14 int y[N];
    15 int c[N];
    16 int MID(int a[],int value,int size)
    17 {
    18     int l=1,r=size;
    19     int mid=(l+r)/2;
    20     while(l<r)//1 3 5 7 9 3
    21     {
    22         if(a[mid]<value&&value<=a[mid+1])return mid+1;
    23         else if(value<=a[mid])r=mid;
    24         else l=mid+1;
    25         mid=(l+r)/2;
    26     }
    27 }
    28 int LIS(int len)
    29 {
    30     int size=1;
    31     int zhong=0;
    32     c[1]=y[1];
    33     for(int i=2;i<=len;i++)//7 6 4 9 11 14 17 3 15 20 4 5 6 7 8 
    34     {//3 9 11 15 17 20
    35         if(y[i]<=c[1])c[1]=y[i];
    36         else if(y[i]>c[size])
    37         c[++size]=y[i];
    38         else 
    39         {
    40         zhong=MID(c,y[i],size);
    41         c[zhong]=y[i];
    42         }
    43     }
    44     return size;
    45 }
    46 int main()
    47 {
    48     int n;
    49     int cnt=1;
    50     while(cin>>n)
    51     {
    52         for(int i=0;i<n;i++)
    53         scanf("%d %d",&node[i].from,&node[i].to);
    54         sort(node,node+n);//sort one side by ascending direction
    55         for(int i=0;i<n;i++)
    56             y[i+1]=node[i].to;
    57         int ans=LIS(n);
    58         cout<<"Case "<<cnt<<':'<<endl;
    59         cout<<"My king, at most "<<ans;
    60         if(ans>1)cout<<" roads can be built."<<endl;
    61         else cout<<" road can be built."<<endl;
    62         cnt++;
    63         cout<<endl;
    64     }
    65     return 0;
    66 }


    It has used the longest increasing subsequence algorithm way,and to get it solved after you have learned it.

    input : with a n,represent n pairs of integers followed.And n pairs of intergers followed

    output:

    Cast i:

    My king, at most n road(s) can be built

    A sample below:

    11
    2 5
    3 6
    4 2
    5 1
    6 9
    7 11
    8 7
    9 12
    10 8
    11 4
    12 10
    Case 1:
    My king, at most 6 road can be built.

    What I don't dare to say is I can't!
  • 相关阅读:
    输入n个整数,输出其中最小的k个
    输出单向链表中倒数第k个结点
    扑克牌大小
    Optional<T> 避免和null检查相关的bug
    筛选、切片、匹配、查找、匹配、归约
    java8 流 中的常用函数式接口
    action 和 controller 单例与多例问题
    Collections.sort Comparator.comparing 冒泡排序 效率对比
    java8 流
    将逗号分割的列,变成多列
  • 原文地址:https://www.cnblogs.com/sytu/p/3916549.html
Copyright © 2011-2022 走看看