SDIBT 3237 Boring Counting( 划分树+二分枚举 )

http://acm.sdibt.edu.cn/JudgeOnline/problem.php?id=3237

Problem H:Boring Counting

Time Limit: 3 Sec  Memory Limit: 128 MB Submit: 8  Solved: 4 [Submit][Status][Discuss]

Description

In this problem you are given a number sequence P consisting of N integer and P_i is the i^th element in the sequence. Now you task is to answer a list of queries, for each query, please tell us among [L, R], how many P_i is not less than A and not greater than B( L<= i <= R). In other words, your task is to count the number of P_i (L <= i <= R,  A <= P_i <= B).

Input

In the first line there is an integer T (1 < T <= 50), indicates the number of test cases.        For each case, the first line contains two numbers N and M (1 <= N, M <= 50000), the size of sequence P, the number of queries. The second line contains N numbers P_i(1 <= P_i <= 10^9), the number sequence P. Then there are M lines, each line contains four number L, R, A, B(1 <= L, R <= n, 1 <= A, B <= 10^9)

Output

For each case, at first output a line ‘Case #c:’, c is the case number start from 1. Then for each query output a line contains the answer.

Sample Input

1
13 5
6 9 5 2 3 6 8 7 3 2 5 1 4
1 13 1 10
1 13 3 6
3 6 3 6
2 8 2 8
1 9 1 9

Sample Output

Case #1:
13
7
3
6
9

HINT

Source

山东省第四届ACM程序设计大赛2013.6.9

【题解】：

　　题目大意：求[L,R]区间内的pi在满足[A<=pi<=B]的个数

　　　一开始想得是：直接用线段树记录区间的最大最小值，当最大最小值满足[A<=pi<=B]时就加上整个区间的长度，可是这样做会超时，因为不满足这个条件的情况太多了，基本上不满足的话每次都要遍历到最底层，所以TLE在所难免。

　　后面觉得还是划分树靠谱，不过比赛的时候也没时间写了，划分树是求区间的第K大值，我们要转换一下，就是A和B在区间的分别是第几大，然后用后者减掉前者就是要求的解，那么拿A来说明一下，A在区间中到底是第几大呢，刚说过划分树是求区间的第K大数是谁的，而我们要求的是知道这个数是谁要求他在区间是第几大，有点绕口哈，这里，我们用到二分枚举，枚举区间第K大的数，然后与A进行比较，如果大于等于A，继续向下枚举，直到等于A的最后一个A为止，对于B同理：

时间复杂度，划分树的时间复杂度为O（n*log（n）） 二分枚举为O(logn)  整体时间复杂度为O(n*log(n)*log(n))

【code】：

/**
Judge Status:Accepted   Memory:13260 KB
Time:2500 ms    Language:C++
Code Lenght:2824 B  Author:cj
*/

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>

#define N 50050
using namespace std;

int sorted[N];   //排序完的数组
int toleft[30][N];   //toleft[i][j]表示第i层从1到k有多少个数分入左边
int tree[30][N];  //表示每层每个位置的值
int n;

void building(int l,int r,int dep)
{
    if(l==r)    return;
    int mid = (l+r)>>1;
    int temp = sorted[mid];
    int i,sum=mid-l+1;    //表示等于中间值而且被分入左边的个数
    for(i=l;i<=r;i++)
    {
        if(tree[dep][i]<temp)   sum--;
    }
    int leftpos = l;
    int rightpos = mid+1;
    for(i=l;i<=r;i++)
    {
        if(tree[dep][i]<temp)  //比中间的数小，分入左边
        {
            tree[dep+1][leftpos++]=tree[dep][i];
        }
        else if(tree[dep][i]==temp&&sum>0)  //等于中间的数值，分入左边，直到sum==0后分到右边
        {
             tree[dep+1][leftpos++]=tree[dep][i];
             sum--;
        }
        else   //右边
        {
            tree[dep+1][rightpos++]=tree[dep][i];
        }
        toleft[dep][i] = toleft[dep][l-1] + leftpos - l;   //从1到i放左边的个数
    }
    building(l,mid,dep+1);
    building(mid+1,r,dep+1);
}

//查询区间第k大的数，[L,R]是大区间，[l,r]是要查询的小区间
int query(int L,int R,int l,int r,int dep,int k)
{
    if(l==r)    return tree[dep][l];
    int mid = (L+R)>>1;
    int cnt = toleft[dep][r] - toleft[dep][l-1]; //[l,r]中位于左边的个数
    if(cnt>=k)
    {
        int newl = L + toleft[dep][l-1] - toleft[dep][L-1]; //L+要查询的区间前被放在左边的个数
        int newr = newl + cnt - 1;  //左端点加上查询区间会被放在左边的个数
        return query(L,mid,newl,newr,dep+1,k);
    }
    else
    {
        int newr = r + (toleft[dep][R] - toleft[dep][r]);
        int newl = newr - (r-l-cnt);
        return query(mid+1,R,newl,newr,dep+1,k-cnt);
    }
}

int bilibili(int L,int R,int l,int r,int a)  //二分枚举
{
    int ans=-1;
    while(l<=r)
    {
        int mid = (l+r)>>1;
        int res = query(1,n,L,R,0,mid);
        if(res>=a)  //直到找到最左边的那个等于a的结果
        {
            r = mid - 1;
            ans = mid;
        }
        else
        {
            l = mid + 1;
        }
    }
    return ans;
}

int bulobulo(int L,int R,int l,int r,int b)
{
    int ans=0;
    while(l<=r)
    {
        int mid = (l+r)>>1;
        int res = query(1,n,L,R,0,mid);
        if(res>b)  //直到找到最后边的大于b的结果
        {
            r = mid - 1;
            ans = mid;
        }
        else
        {
            l = mid + 1;
        }
    }
    if(!ans)    return r;
    return ans-1;
}


int main()
{
    int t,cas = 1;
    scanf("%d",&t);
    while(t--)
    {
        int m;
        scanf("%d%d",&n,&m);
        int i;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&tree[0][i]);
            sorted[i] = tree[0][i];
        }
        sort(sorted+1,sorted+1+n);
        building(1,n,0);
        int l,r,a,b;
        printf("Case #%d:
",cas++);
        while(m--)
        {
            scanf("%d%d%d%d",&l,&r,&a,&b);
            int x = 1;
            int y = r-l+1;
            int cnt1 = bilibili(l,r,x,y,a);
            int cnt2 = bulobulo(l,r,x,y,b);
            if(cnt1==-1)
            {
                printf("0
");
                continue;
            }
            printf("%d
",cnt2-cnt1+1);
        }
    }
    return 0;
}