【山东省第四节ACM省赛】Boring Counting（二分+划分树）

算是一道模板题了，可惜弱的一B的我并不会划分树，花了点时间学了下，回头A了这道题

3s的限时跑了2.8s也是醉了。。。

Description

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).

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 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)

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:

1

3

7

3

6

9

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn = 50005;
int n,m;
int tree[20][maxn];
int lnum[20][maxn];
int sorted[maxn];
void debug(int v){
    for(int i = 1; i <= n; i++) printf("%d ",sorted[i]);
    puts("");
    for(int i = 0; i < v; i++){
        for(int j = 1; j <= n; j++)
            printf("%d ",tree[i][j]);
        puts("");
    }
}
void build(int l,int r,int d){
    if(l == r) return;
    int mid = (l + r) >> 1;
    int same = mid - l + 1;
    for(int i = l; i <= r; i++)
        if(tree[d][i] < sorted[mid])
            same --;
    int lpos = l,rpos = mid + 1;
    lnum[d][0] = 0;
    for(int i = l; i <= r; i++){
        if(tree[d][i] < sorted[mid]){
            lnum[d][i] = lnum[d][i - 1] + 1;
            tree[d + 1][lpos++] = tree[d][i];
        }
        else if(tree[d][i] == sorted[mid] && same > 0){
            same --;
            lnum[d][i] = lnum[d][i - 1] + 1;
            tree[d + 1][lpos++] = tree[d][i];
        }
        else{
            lnum[d][i] = lnum[d][i - 1];
            tree[d + 1][rpos++] = tree[d][i];
        }
    }
    build(l,mid,d + 1);
    build(mid + 1,r,d + 1);
}
int query(int L,int R,int l,int r,int d,int k){
    //printf("%d %d %d %d\n",l,r,d,k);
    if(l == r) return tree[d][l];
    int mid = (L + R) >> 1;
    int cnt = lnum[d][r] - lnum[d][l - 1];
    if(cnt >= k){
        int newl = L + lnum[d][l - 1] - lnum[d][L - 1];
        int newr = newl + cnt - 1;
        return query(L,mid,newl,newr,d + 1,k);
    }
    else{
        int newr = r + lnum[d][R] - lnum[d][r];
        int newl = newr - (r - l - cnt);
        return query(mid + 1,R,newl,newr,d + 1,k - cnt);
    }
}
void solve(){
    int L,R,a,b;
    scanf("%d%d%d%d",&L,&R,&a,&b);
    int l = 1,r = (R - L + 2);
    int v1 = 0,v2 = 0;
    //找大于等于
    while(l < r){
        int mid = (l + r) >> 1;
        int value = query(1,n,L,R,0,mid);
        if(value < a)
            l = mid + 1;
        else if(value >= a)
            r = mid;
    }
    v1 = l;
    l = 1;
    r = (R - L + 2);
    //找大于
    while(l < r){
        int mid = (l + r) >> 1;
        int value = query(1,n,L,R,0,mid);
        if(value <= b){
            l = mid + 1;
        }
        else if(value > b)
            r = mid;
    }
    v2 = l;
    printf("%d\n",v2 - v1);
}
int main(){
    int T,Case = 1;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&n,&m);
        for(int i = 1; i <= n; i++){
            scanf("%d",&tree[0][i]);
            sorted[i] = tree[0][i];
        }
        sort(sorted + 1,sorted + n + 1);
        build(1,n,0);
        printf("Case #%d:\n",Case++);
        for(int i = 0; i < m; i++)
            solve();
    }
    return 0;
}
/*
5 5
1 1 1 1 1
1 5 2 3
2
*/