http://acm.sdibt.edu.cn/JudgeOnline/problem.php?id=3237
Problem H:Boring Counting
Time Limit: 3 Sec Memory Limit: 128 MB Submit: 8 Solved: 4 [Submit][Status][Discuss]
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:
13
7
3
6
9
HINT
Source
【题解】：
题目大意：求[L,R]区间内的pi在满足[A<=pi<=B]的个数
一开始想得是：直接用线段树记录区间的最大最小值，当最大最小值满足[A<=pi<=B]时就加上整个区间的长度，可是这样做会超时，因为不满足这个条件的情况太多了，基本上不满足的话每次都要遍历到最底层，所以TLE在所难免。
后面觉得还是划分树靠谱，不过比赛的时候也没时间写了，划分树是求区间的第K大值，我们要转换一下，就是A和B在区间的分别是第几大，然后用后者减掉前者就是要求的解，那么拿A来说明一下，A在区间中到底是第几大呢，刚说过划分树是求区间的第K大数是谁的，而我们要求的是知道这个数是谁要求他在区间是第几大，有点绕口哈，这里，我们用到二分枚举，枚举区间第K大的数，然后与A进行比较，如果大于等于A，继续向下枚举，直到等于A的最后一个A为止，对于B同理：
时间复杂度，划分树的时间复杂度为O（n*log（n）） 二分枚举为O(logn) 整体时间复杂度为O(n*log(n)*log(n))
【code】：
/**
Judge Status:Accepted Memory:13260 KB
Time:2500 ms Language:C++
Code Lenght:2824 B Author:cj
*/ #include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm> #define N 50050
using namespace std; int sorted[N]; //排序完的数组
int toleft[][N]; //toleft[i][j]表示第i层从1到k有多少个数分入左边
int tree[][N]; //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)>>;
int temp = sorted[mid];
int i,sum=mid-l+; //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+;
for(i=l;i<=r;i++)
{
if(tree[dep][i]<temp) //比中间的数小，分入左边
{
tree[dep+][leftpos++]=tree[dep][i];
}
else if(tree[dep][i]==temp&&sum>) //等于中间的数值，分入左边，直到sum==0后分到右边
{
tree[dep+][leftpos++]=tree[dep][i];
sum--;
}
else //右边
{
tree[dep+][rightpos++]=tree[dep][i];
}
toleft[dep][i] = toleft[dep][l-] + leftpos - l; //从1到i放左边的个数
}
building(l,mid,dep+);
building(mid+,r,dep+);
} //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)>>;
int cnt = toleft[dep][r] - toleft[dep][l-]; //[l,r]中位于左边的个数
if(cnt>=k)
{
int newl = L + toleft[dep][l-] - toleft[dep][L-]; //L+要查询的区间前被放在左边的个数
int newr = newl + cnt - ; //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+,k);
}
else
{
int newr = r + (toleft[dep][R] - toleft[dep][r]);
int newl = newr - (r-l-cnt);
return query(mid+,R,newl,newr,dep+,k-cnt);
}
} int bilibili(int L,int R,int l,int r,int a) //二分枚举
{
int ans=-;
while(l<=r)
{
int mid = (l+r)>>;
int res = query(,n,L,R,,mid);
if(res>=a) //直到找到最左边的那个等于a的结果
{
r = mid - ;
ans = mid;
}
else
{
l = mid + ;
}
}
return ans;
} int bulobulo(int L,int R,int l,int r,int b)
{
int ans=;
while(l<=r)
{
int mid = (l+r)>>;
int res = query(,n,L,R,,mid);
if(res>b) //直到找到最后边的大于b的结果
{
r = mid - ;
ans = mid;
}
else
{
l = mid + ;
}
}
if(!ans) return r;
return ans-;
} int main()
{
int t,cas = ;
scanf("%d",&t);
while(t--)
{
int m;
scanf("%d%d",&n,&m);
int i;
for(i=;i<=n;i++)
{
scanf("%d",&tree[][i]);
sorted[i] = tree[][i];
}
sort(sorted+,sorted++n);
building(,n,);
int l,r,a,b;
printf("Case #%d:\n",cas++);
while(m--)
{
scanf("%d%d%d%d",&l,&r,&a,&b);
int x = ;
int y = r-l+;
int cnt1 = bilibili(l,r,x,y,a);
int cnt2 = bulobulo(l,r,x,y,b);
if(cnt1==-)
{
printf("0\n");
continue;
}
printf("%d\n",cnt2-cnt1+);
}
}
return ;
}