山東省第四屆ACM大學生程式設計競賽-Boring Counting（劃分樹-二分查詢）

kewlgrl發表於2016-05-07
原文網址 : https://blog.csdn.net/mikasa3/article/details/51339948
ACM程式設計

Boring Counting

Time Limit: 3000ms   Memory limit: 65536K  有疑問？點這裡^_^

題目描述

    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).

輸入

     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)

輸出

    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.

示例輸入

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

示例輸出

Case #1:
13
7
3
6
9

提示

 

來源

 2013年山東省第四屆ACM大學生程式設計競賽

題目意思：

求每組數中，在l~r的範圍內，有多少個數的值在a~b的範圍內。

解題思路：

劃分樹+二分查詢。
之前寫過遍歷查詢直接超時……用二分的效率很高。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define N 50050
using namespace std;

int sorted[N];   //排序完的陣列
int toleft[30][N];   //toleft[i][j]表示第i層從1到k有多少個數分入左邊
int tree[30][N];  //表示每層每個位置的值
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;    //表示等於中間值而且被分入左邊的個數
    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)  //比中間的數小，分入左邊
            tree[dep+1][leftpos++]=tree[dep][i];
        else if(tree[dep][i]==temp&&sum>0)  //等於中間的數值，分入左邊，直到sum==0後分到右邊
        {
            tree[dep+1][leftpos++]=tree[dep][i];
            sum--;
        }
        else   //右邊
            tree[dep+1][rightpos++]=tree[dep][i];
        toleft[dep][i] = toleft[dep][l-1] + leftpos - l;   //從1到i放左邊的個數
    }
    building(l,mid,dep+1);
    building(mid+1,r,dep+1);
}

//查詢區間第k大的數，[L,R]是大區間，[l,r]是要查詢的小區間
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]中位於左邊的個數
    if(cnt>=k)
    {
        int newl = L + toleft[dep][l-1] - toleft[dep][L-1]; //L+要查詢的區間前被放在左邊的個數
        int newr = newl + cnt - 1;  //左端點加上查詢區間會被放在左邊的個數
        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 MAXA(int L,int R,int l,int r,int a)  //二分列舉
{
    int ans=-1;
    while(l<=r)
    {
        int mid = (l+r)>>1;
        int res = query(1,n,L,R,0,mid);
        if(res>=a)  //直到找到最左邊的那個等於a的結果
        {
            r = mid - 1;
            ans = mid;
        }
        else l = mid + 1;
    }
    return ans;
}

int MINB(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)  //直到找到最後邊的大於b的結果
        {
            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:\n",cas++);
        while(m--)
        {
            scanf("%d%d%d%d",&l,&r,&a,&b);
            int x = 1;
            int y = r-l+1;
            int cnt1 = MAXA(l,r,x,y,a);
            int cnt2 = MINB(l,r,x,y,b);
            if(cnt1==-1)
            {
                printf("0\n");
                continue;
            }
            printf("%d\n",cnt2-cnt1+1);
        }
    }
    return 0;
}
 



/**************************************
	Problem id	: SDUT OJ 2610 
	User name	: 吃花的栗鼠 
	Result		: Accepted 
	Take Memory	: 7260K 
	Take Time	: 1230MS 
	Submit Time	: 2016-05-07 20:27:46  
**************************************/


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