sdut2610--Boring Counting(二分+划分树)

Boring Counting

Time Limit: 3000ms   Memory limit: 65536K  有疑问？点这里^_^

题目描述

    In this problem you are given a number sequence P consisting of N integer and Pi is the ith 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 Pi is not less than A and not greater than B( L<= i <= R). In other words, your task is to count the number of Pi (L <= i <= R,  A <= Pi <= B).

输入

     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 Pi(1 <= Pi <= 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)

输出

    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.

示例输入

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

示例输出

Case #1:
13
7
3
6
9

 

给出n个数，问范围在[l,r]内，值在[A,B]内的数有多少个

转化为在[l,r]的范围内A是第几大的数，B是第几大的树，这样相减就可以得到结果，但是我们只能求第k大的数，不能找出数是第几个，二分k，用二分找出上下界。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define maxn 50005
int a[maxn] , a_sort[maxn] ;
int tree[20][maxn] ;
int sum[20][maxn] ;
int n , m ;
void build(int c,int l,int r)
{
    if( l == r ) return ;
    int mid = (l+r)/2 , m = mid-l+1 ;
    int pl = l , pr = mid+1 , i ;
    for(i = l ; i <= mid ; i++)
    {
        if( a_sort[i] < a_sort[mid] )
            m-- ;
    }
    for(i = l ; i <= r ; i++)
    {
        if( i == l ) sum[c][i] = 0 ;
        else sum[c][i] = sum[c][i-1] ;
        if( tree[c][i] == a_sort[mid] )
        {
            if( m )
            {
                m-- ;
                tree[c+1][pl++] = tree[c][i] ;
                sum[c][i]++ ;
            }
            else
                tree[c+1][pr++] = tree[c][i] ;
        }
        else if( tree[c][i] < a_sort[mid] )
        {
            tree[c+1][pl++] = tree[c][i] ;
            sum[c][i]++ ;
        }
        else
            tree[c+1][pr++] = tree[c][i] ;
    }
    build(c+1,l,mid) ;
    build(c+1,mid+1,r) ;
    return ;
}
int query(int c,int l,int r,int ql,int qr,int k) {
    if( l == r ) return tree[c][l] ;
    int mid = (l+r)/2 , num1 , num2 ;
    if( l == ql ) {
        num1 = 0 ;
        num2 = sum[c][qr] ;
    }
    else {
        num1  = sum[c][ql-1] ;
        num2 = sum[c][qr] - num1 ;
    }
    if( k <= num2 )
        query(c+1,l,mid,l+num1,l+num1+num2-1,k) ;
    else
        query(c+1,mid+1,r,mid+1+(ql-l-num1),mid+1+(qr-l-num1-num2),k-num2) ;
}
void solve(int l,int r,int A,int B) {
    int low , mid , high , last , x ;
    int ans = 0 , flag = 0 ;
    low = 1 ; high = r-l+1 ; last = -1 ;
    while( low <= high ) {
        mid = (low+high)/2 ;
        x = query(0,1,n,l,r,mid) ;
        if( x <= B ) {
            last = mid ;
            low = mid + 1 ;
        }
        else
            high = mid - 1 ;
    }
    if( last == -1 )
        flag = 1 ;
    else
        ans += last ;
    low = 1 ; high = r-l+1 ; last = -1 ;
    while( low <= high ) {
        mid = (low+high)/2 ;
        x = query(0,1,n,l,r,mid) ;
        if( x >= A ) {
            last = mid ;
            high = mid - 1 ;
        }
        else
            low = mid + 1 ;
    }
    if( last == -1 )
        flag = 1 ;
    else
        ans = ans - last+1 ;
    if( flag ) ans = 0 ;
    printf("%d\n", ans) ;
}
int main()
{
    int t , step = 0 ;
    int l , r , A , B ;
    int i ;
    scanf("%d", &t) ;
    while( t-- )
    {
        scanf("%d %d", &n, &m) ;
        for(i = 1 ; i <= n ; i++)
        {
            scanf("%d", &a[i]) ;
            tree[0][i] = a_sort[i] = a[i] ;
        }
        sort(a_sort+1,a_sort+1+n) ;
        build(0,1,n) ;
        printf("Case #%d:\n", ++step) ;
        while( m-- ) {
            scanf("%d %d %d %d", &l, &r, &A, &B) ;
            solve(l,r,A,B) ;
        }
    }
    return 0 ;
}