KPI

6
in 874
query
out
in 24622
in 12194
query

Case #1:
874
24622

Description

Input

Output

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.

这两题都 要求子区间的第k大数，假设用快排之后，再查，复杂度太大。不能过，就要用到划分树了，划分树。事实上，就是线段树的

一个变种了，当然，这样做查询是lg(n)级别，建树是n * lg(n),再加上一次快排，也是n * lg(n),差点儿相同就能够过了！

先来看看划分树，我们在线段树的基础上，假设每一个结点都保存了。当前子段向左子树走的个数。假设，查询的k要小于，查询段向左走的个数，自然，我们向左子树就能够查到结果。假设，k>向左走的个数，当然要向右找k-midcount个数，区间的变换。我们能够

推一下向左走就是l + scount, l + ecount - 1。向右走就是mid + 1 + s - l - scount, mid + 1 + e - l - ecount，当中scount就是s之前向左走的个数。 ecount，就是e之前向左走的个数。

上核心代码

#define MID(a,b) (((a)+(b))>>1)
#define N 100050
int num[20][N], val[20][N];
//第i层，向左包含自已，向左子树的个数，当前的值
int pri[N], sorted[N];
//原始和排序后的数列
char str[20];
void build(int l, int r, int layer){
	if (l >= r){
		return;
	}
	int mid = MID(l, r);
	int ql = l, qr = mid + 1, leftCount = mid, eqCount = 0;
	for (int i = l; i <= r; i++)
		leftCount -= val[layer][i] < sorted[mid];
	for (int i = l; i <= r; i++){
		if (i == l){
			num[layer][i] = 0;
		}
		else{
			num[layer][i] = num[layer][i - 1];
		}
		if (val[layer][i] < sorted[mid]){
			num[layer][i]++;
			val[layer + 1][ql++] = val[layer][i];
		}
		else if (val[layer][i] > sorted[mid]){
			val[layer + 1][qr++] = val[layer][i];
		}
		else {
			if (eqCount < leftCount){
				eqCount++;
				num[layer][i]++;
				val[layer + 1][ql++] = val[layer][i];
			}
			else {
				val[layer + 1][qr++] = val[layer][i];
			}
		}
	}
	build(l, mid, layer + 1);
	build(mid + 1, r, layer + 1);
}
//在layer层l到r间，找s-e之间的第k大数
int query(int l, int r, int layer, int s, int e, int k){
	if (l >= r || s >= e){
		return val[layer][s];
	}
	int scount, ecount, mid = MID(l, r);
	if (s == l){
		scount = 0, ecount = num[layer][e];
	}
	else {
		scount = num[layer][s - 1], ecount = num[layer][e];
	}
	int midcount = ecount - scount;
	if (k <= midcount){
		return query(l, mid, layer + 1, l + scount, l + ecount - 1, k);
	}
	else {
		return query(mid + 1, r, layer + 1, mid + 1 + s - l - scount, mid + 1 + e - l - ecount, k - midcount);
	}
}
int main()
{
	int tcase, tcasenum = 0;
	int n, k, q, s, e, begin = 1, nn, qcount, m;
	while (S2(n, m) != EOF)
	{
		FI(n){
			S(val[0][i+1]);
		}
		FI(n){
			sorted[i + 1] = val[0][i + 1];
		}
		sort(sorted + 1, sorted + n + 1);
		build(1, n, 0);
		FI(m){
			S2(s, e); S(q);
			Prn(query(1, n, 0, s,e,q));
		}
	}
	return 0;
}