【HDU】4417 Super Mario（划分树+二分）

这道题目是利用了划分树，通过枚举这个区间的第几小的数是否小于k来找出答案。

而枚举的方法则是二分。

有大神说可以不用二分，不知道是怎么搞的，求指教

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define MAX 100050
#define m (l+r)>>1
int tree[30][MAX];
int toleft[30][MAX];
int sorted[MAX];
int n, k;
void build(int l, int r, int dep)
{
	if (l == r)return;
	int mid = m;
	int cnt = mid - l + 1;
	for (int i = l; i <= r;i++)
	if (tree[dep][i] < sorted[mid])
		cnt--;
	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] && cnt)
		{
			tree[dep + 1][lpos++] = tree[dep][i];
			cnt--;
		}
		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);
}

int query(int L, int R, int l, int r, int dep, int k)
{
	if (l == r)return tree[dep][l];
	int cnt = toleft[dep][R] - toleft[dep][L - 1];
	int mid = m;
	if (cnt >= k)
	{
		int newL = l + toleft[dep][L - 1] - toleft[dep][l - 1];
		int newR = newL + cnt - 1;
		return query(newL, newR, l, mid, dep + 1, k);
	}
	else
	{
		int newR = R + toleft[dep][r] - toleft[dep][R];
		int newL = newR - (R - L - cnt);
		return query(newL, newR, mid + 1, r, dep + 1, k - cnt);
	}
}

int gao(int s, int t,int h)
{
	int mid=0;
	int r =t - s + 1;
	int l = 1;
	int ans = 0;
	while (l <= r)
	{
		mid = m;
		int temp = query(s, t, 1, n, 0, mid);
		if (temp <= h)
		{
			ans = mid;
			l = mid + 1;

		}
		else
			r = mid - 1;
	}
	return ans;
}

int main()
{
	int t;
	scanf("%d", &t);
	int icase = 1;
	while (t--)
	{
		scanf("%d%d", &n, &k);
		for (int i = 1; i <=n; i++)
		{
			scanf("%d", &tree[0][i]);
			sorted[i] = tree[0][i];
		}
		sort(sorted+1, sorted+1 + n);
		build(1, n, 0);
		int a, b, c;
		printf("Case %d:\n", icase++);
		while (k--)
		{
			scanf("%d%d%d", &a, &b, &c);
			printf("%d\n", gao(a+1, b+1, c));//由于上面我是从1开始，而题目是从0开始。所以要进1
		}
	}
}