poj 2761 Feed the dogs 划分树

最新推荐文章于 2018-12-31 15:23:36 发布

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本文介绍了一种解决喂狗问题的独特算法，该问题涉及在不重复覆盖的情况下选择特定美观度的狗进行喂食。通过构建多层次的数据结构，算法能够高效地确定每次喂食的对象，避免了对同一位置的重复喂食，确保了所有狗都能获得食物。

Feed the dogs

Time Limit: 6000MS		Memory Limit: 65536K
Total Submissions: 22814		Accepted: 7233

Description

Wind loves pretty dogs very much, and she has n pet dogs. So Jiajia has to feed the dogs every day for Wind. Jiajia loves Wind, but not the dogs, so Jiajia use a special way to feed the dogs. At lunchtime, the dogs will stand on one line, numbered from 1 to n, the leftmost one is 1, the second one is 2, and so on. In each feeding, Jiajia choose an inteval[i,j], select the k-th pretty dog to feed. Of course Jiajia has his own way of deciding the pretty value of each dog. It should be noted that Jiajia do not want to feed any position too much, because it may cause some death of dogs. If so, Wind will be angry and the aftereffect will be serious. Hence any feeding inteval will not contain another completely, though the intervals may intersect with each other.

Your task is to help Jiajia calculate which dog ate the food after each feeding.

Input

The first line contains n and m, indicates the number of dogs and the number of feedings.

The second line contains n integers, describe the pretty value of each dog from left to right. You should notice that the dog with lower pretty value is prettier.

Each of following m lines contain three integer i,j,k, it means that Jiajia feed the k-th pretty dog in this feeding.

You can assume that n<100001 and m<50001.

Output

Output file has m lines. The i-th line should contain the pretty value of the dog who got the food in the i-th feeding.

Sample Input

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

Sample Output

3
2

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<cctype>
using namespace std;
const int MAXN = 100010;
int tree[20][MAXN];//表示每层每个位置的值
int sorted[MAXN];//已经排序好的数
int toleft[20][MAXN];//toleft[p][i] 表示第 i 层从 1 到 i 有数分入左边

void build(int l,int r,int dep) {
	if(l == r)return;
	int mid = (l+r)>>1;
	int same=mid-l+1;//表示等于中间值而且被分入左边的个数
	for(int i = l; i <= r; i++) //注意是 l, 不是 one
		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);
}

//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];
	if(cnt >= k) {
        int newl=L+toleft[dep][l-1]-toleft[dep][L-1];

		int newr=newl+cnt-1;

		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 main() {
	int n,m;
	while(scanf("%d%d",&n,&m)==2) {
		memset(tree,0,sizeof(tree));
		for(int i = 1; i <= n; i++) {
			scanf("%d",&tree[0][i]);
			sorted[i] = tree[0][i];
		}
		sort(sorted+1,sorted+n+1);
		build(1,n,0);
		int s,t,k;
		while(m--) {
			scanf("%d%d%d",&s,&t,&k);
			printf("%d\n",query(1,n,s,t,0,k));
		}
	}
	return 0;
}