山东省第四届ACM大学生程序设计竞赛-Boring Counting（划分树-二分查找）

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

提示

 

来源

 2013年山东省第四届ACM大学生程序设计竞赛

题目意思：

求每组数中，在l~r的范围内，有多少个数的值在a~b的范围内。

解题思路：

划分树+二分查找。

之前写过遍历查找直接超时……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
;   //排序完的数组
int toleft[30]
;   //toleft[i][j]表示第i层从1到k有多少个数分入左边
int tree[30]
;  //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 MAXA(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 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:\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
**************************************/