• ZOJ 3633 Alice's present【线段树】


    As a doll master, Alice owns a wide range of dolls, and each of them has a number tip on it's back, the tip can be treated as a positive integer. (the number can be repeated). One day, Alice hears that her best friend Marisa's birthday is coming , so she decides to sent Marisa some dolls for present. Alice puts her dolls in a row and marks them from 1 to n. Each time Alice chooses an interval from i to j in the sequence ( include i and j ) , and then checks the number tips on dolls in the interval from right to left. If any number appears more than once , Alice will treat this interval as unsuitable. Otherwise, this interval will be treated as suitable.

    This work is so boring and it will waste Alice a lot of time. So Alice asks you for help .

    Input

    There are multiple test cases. For each test case:

    The first line contains an integer n ( 3≤ n ≤ 500,000) ,indicate the number of dolls which Alice owns.

    The second line contains n positive integers , decribe the number tips on dolls. All of them are less than 2^31-1. The third line contains an interger m ( 1 ≤ m ≤ 50,000 ),indicate how many intervals Alice will query. Then followed by m lines, each line contains two integer u, v ( 1≤ u< vn ),indicate the left endpoint and right endpoint of the interval. Process to the end of input.

    Output

    For each test case:

    For each query, If this interval is suitable , print one line "OK". Otherwise, print one line ,the integer which appears more than once first.

    Print an blank line after each case.

    Sample Input

    5
    1 2 3 1 2
    3
    1 4
    1 5
    3 5
    6
    1 2 3 3 2 1
    4
    1 4
    2 5
    3 6
    4 6
    

    Sample Output

    1
    2
    OK
    
    3
    3
    3
    OK
    

    题目大意:给出n个数,在区间[x, y]内找出从右向左出现不止一次的数,输出最右端的数字;没有输出OK;
    思路:区间查询,就该往线段树方向上想;那么当百每个数字的左端离他最近的数字就录下来,在区间【x, y】内寻找最大值,那么就是右端起最先重复出现的数字,但是必须要    满足最大的数maxnum>=x, 否则是属于没有重复数字的,输出ok;

    代码如下:

    View Code
    #include<stdio.h>
    #include<string.h>
    #include<iostream>
    #include<map>
    #include<math.h> 
    #include<algorithm>
    using namespace std;
    #define N  500005
    int dpmax[N][19], a[N]; 
    int main()
    {
        int i, j, n, m, x, y;
        while(scanf("%d", &n)!=EOF)
        {
            map<int, int>M; 
            memset(a, 0, sizeof(a)); 
            for(i=1; i<=n; i++)
            {
                scanf("%d", &a[i]);
                dpmax[i][0]=M[a[i]];
                M[a[i]]=i;
            } 
            int mm=(int)floor(log(1.0*n)/log(2.0)); //n==500000时, mm=18 
            for(j=1; j<=mm; j++)
                for(i=n; i>=1; i--)
                    if((i+(1<<(j-1)))<=n)
                        dpmax[i][j]=max(dpmax[i][j-1], dpmax[i+(1<<(j-1))][j-1]);
             scanf("%d", &m);
            for(i=1; i<=m; i++)
            {
                scanf("%d%d", &x, &y);
                int  mid=(int)floor(log(y*1.0-x+1)/log(2.0));
                int maxnum=max(dpmax[x][mid], dpmax[y-(1<<mid)+1][mid]);
                if(maxnum<x)
                    printf("OK\n");
                else
                    printf("%d\n",  a[maxnum]);
            }
            printf("\n"); 
        }
        return 0;
    }             


  • 相关阅读:
    递归方法:对于树形结构的表,根据当前数据获取无限极的父级名称
    P
    A
    今年暑假不AC1
    J
    今年暑假不AC
    A
    *max_element函数和*min_element函数
    1199: 房间安排
    素数
  • 原文地址:https://www.cnblogs.com/Hilda/p/2662574.html
Copyright © 2020-2023  润新知