Super Mario

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4041    Accepted Submission(s): 1862


Problem Description Mario is world-famous plumber. His “burly” figure and amazing jumping ability reminded in our memory. Now the poor princess is in trouble again and Mario needs to save his lover. We regard the road to the boss’s castle as a line (the length is n), on every integer point i there is a brick on height hi. Now the question is how many bricks in [L, R] Mario can hit if the maximal height he can jump is H.
Input The first line follows an integer T, the number of test data.
For each test data:
The first line contains two integers n, m (1 <= n <=10^5, 1 <= m <= 10^5), n is the length of the road, m is the number of queries.
Next line contains n integers, the height of each brick, the range is [0, 1000000000].
Next m lines, each line contains three integers L, R,H.( 0 <= L <= R < n 0 <= H <= 1000000000.)
Output For each case, output "Case X: " (X is the case number starting from 1) followed by m lines, each line contains an integer. The ith integer is the number of bricks Mario can hit for the ith query.

Sample Input 1 10 10 0 5 2 7 5 4 3 8 7 7 2 8 6 3 5 0 1 3 1 1 9 4 0 1 0 3 5 5 5 5 1 4 6 3 1 5 7 5 7 3
Sample Output Case 1: 4 0 0 3 1 2 0 1 5 1

題意：

查詢指定區間內比x小的數的個數

思路：

稍微修改下劃分樹，但是不知道為什麼一開始想的是二分- -，強行GG。寫著很麻煩而且還是錯的

劃分數能很方便地求出了區間中小於等於第i個數的個數，在這個基礎上稍作修改即可

/*
*/

#include <functional>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <Map>
using namespace std;
typedef long long ll;
typedef long double ld;

using namespace std;

const int maxn = 100010;

int tree[20][maxn];
int sorted[maxn];
int toleft[20][maxn];

void build(int l,int r,int dep)  //模擬快排 並記錄左樹中比i小的個數
{
    if(l == r)
        return;
    int mid = (l+r)>>1;
    int same = mid-l+1;
    for(int i = l; i <= r; i++)
    {
        if(tree[dep][i] < sorted[mid])
            same--;
    }
    int lpos = l;
    int rpos = mid+1;
    for(int i = l; i <= r; i++)
    {
        if(tree[dep][i] < sorted[mid])
        {
            tree[dep+1][lpos++] = tree[dep][i];
        }
        else if(tree[dep][i] == sorted[mid] && same > 0)
        {
            tree[dep+1][lpos++] = tree[dep][i];
            same --;
        }
        else
        {
            tree[dep+1][rpos++] = tree[dep][i];
        }
        toleft[dep][i] = toleft[dep][l-1] + lpos -l;

    }
    build(l,mid,dep+1);
    build(mid+1,r,dep+1);
}


int query(int L,int R,int l,int r,int dep,int k)
{
    if(l == r)
    {
        if(tree[dep][l] <= k)
            return 1;
        return 0;
    }
    int mid = (L+R)>>1;
    int cnt = toleft[dep][r]-toleft[dep][l-1];
    if(sorted[mid] > k)          //如果比分界線小
    {
        int lpos = L+toleft[dep][l-1]-toleft[dep][L-1];
        int rpos = lpos+cnt-1;
        if(rpos >= lpos)
            return query(L,mid,lpos,rpos,dep+1,k);
        return 0;
    }
    else
    {
        int rpos = r+toleft[dep][R]-toleft[dep][r];
        int lpos = rpos-(r-l-cnt);
        return cnt+query(mid+1,R,lpos,rpos,dep+1,k);
    }
}


int main()
{
    int n,m,T;
    int cas = 1;
    scanf("%d",&T);
    while(T--)
    {

        scanf("%d%d",&n,&m);
        for(int i = 1; i <= n; i++)
        {
            scanf("%d",&sorted[i]);
            tree[0][i] = sorted[i];
        }
        sort(sorted+1,sorted+n+1);
        build(1,n,0);

        int l,r,to,ans;
        printf("Case %d:\n",cas++);
        while(m--)
        {
            scanf("%d%d%d",&l,&r,&to);
            l++;
            r++;
            ans = query(1,n,l,r,0,to);
            printf("%d\n",ans);
        }
    }
    return 0;
}