POJ2104 K-th Number —— 区间第k小 整体二分

题目链接：https://vjudge.net/problem/POJ-2104

 

K-th Number

Time Limit: 20000MS		Memory Limit: 65536K
Total Submissions: 64110		Accepted: 22556
Case Time Limit: 2000MS

Description

You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment. 
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?" 
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.

Input

The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000). 
The second line contains n different integer numbers not exceeding 10⁹ by their absolute values --- the array for which the answers should be given. 
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).

Output

For each question output the answer to it --- the k-th number in sorted a[i...j] segment.

Sample Input

7 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3

Sample Output

5
6
3

Hint

This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.

Source

Northeastern Europe 2004, Northern Subregion

 

题解：

查询区间第k小（不带修改），整体二分。

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 1e9+7;
const int MAXN = 2e5+100;

struct node
{
    int x, y, k, type, id;
};
node q[MAXN];

int n, m, c[MAXN];
int lowbit(int x) {return x&(-x);}
void add(int x, int val) {for(int i=x;i<=n;i+=lowbit(i)) c[i]+=val;}
int sum(int x) {int ret=0; for(int i=x;i>0;i-=lowbit(i))ret+=c[i]; return ret;}

int ans[MAXN];
node q1[MAXN], q2[MAXN];    //两个桶
void solve(int l, int r, int ql, int qr)    //二分答案
{
    if(ql>qr) return;   //！！
    if(l==r)        //当l==r时，即答案已明确
    {
        for(int i = ql; i<=qr; i++)
            if(q[i].type==1) ans[q[i].id] = l;
        return;
    }

    int mid = (l+r)>>1;     //写成 (l+r)/2会runtime error，不知为何
    int t1 = 0, t2 = 0;
    for(int i = ql; i<=qr; i++)     //枚举操作
    {
        if(q[i].type==0)
        {
            if(q[i].y<=mid)
            {
                add(q[i].x, 1);
                q1[++t1] = q[i];
            }else q2[++t2] = q[i];
        }
        else
        {
            int pre = sum(q[i].y)-sum(q[i].x-1);
            if(pre>=q[i].k) q1[++t1] = q[i];
            else
            {
                q[i].k -= pre;
                q2[++t2] = q[i];
            }
        }
    }
    for(int i = 1; i<=t1; i++)  //撤回对线段树的操作
        if(q1[i].type==0) add(q1[i].x, -1);

    for(int i = 1; i<=t1; i++) q[ql+i-1] = q1[i];
    for(int i = 1; i<=t2; i++) q[ql+t1+i-1] = q2[i];
    solve(l, mid, ql, ql+t1-1);
    solve(mid+1, r, ql+t1, qr);
}

int main()
{
    while(scanf("%d%d",&n,&m)==2)
    {
        int tot = 0;
        for(int i = 1; i<=n; i++)
        {
            ++tot;
            scanf("%d", &q[tot].y);
            q[tot].x = i; q[tot].type = 0;
        }
        for(int i = 1; i<=m; i++)
        {
            ++tot;
            scanf("%d%d%d", &q[tot].x,&q[tot].y,&q[tot].k);
            q[tot].id = i; q[tot].type = 1;
        }

        memset(c, 0, sizeof(c));
        solve(-MOD,MOD, 1,tot);
        for(int i = 1; i<=m; i++)
            printf("%d\n", ans[i]);
    }
}