排序（线段树合并与分裂板题）

题目链接：jzoj 4605 / luogu P2824

题目大意

给你一个全排列，然后每次操作会将数列中的某个区间按升序或降序排序。
然后问你最后第 i 个位置上的数是什么。

思路

我们考虑二分，二分这个答案，然后大于等于它的是 1 1 1，小于的是 0 0 0。
那就变成看这个位置是 1 1 1 是 0 0 0，排序就先找到这个区间 0 , 1 0,1 0,1 的个数，然后再按要求区间修改放下去，这个用线段树搞就可以了。

然后二分出来答案就可以了。

代码

#include<cstdio>

using namespace std;

struct ask {
	int op, l, r;
}q[200001];
int n, m, a[200001], pl;
int ans, l, r, mid, nm0, nm1;
bool s[200001];
struct Tree {
	int num0, num1;
	bool all0, all1;
}t[800001];

void up(int now) {//标记上传下传
	t[now].num0 = t[now << 1].num0 + t[now << 1 | 1].num0;
	t[now].num1 = t[now << 1].num1 + t[now << 1 | 1].num1;
	if (t[now << 1].all0 && t[now << 1 | 1].all0) t[now].all0 = 1;
		else t[now].all0 = 0;
	if (t[now << 1].all1 && t[now << 1 | 1].all1) t[now].all1 = 1;
		else t[now].all1 = 0;
}

void down(int now, int l, int r) {
	int mid = (l + r) >> 1;
	if (t[now].all0) {
		t[now << 1].all0 = t[now << 1 | 1].all0 = 1;
		t[now << 1].num0 = mid - l + 1;
		t[now << 1 | 1].num0 = t[now].num0 - t[now << 1].num0;
		t[now << 1].all1 = t[now << 1 | 1].all1 = 0;
		t[now << 1].num1 = t[now << 1 | 1].num1 = 0;
	}
	if (t[now].all1) {
		t[now << 1].all1 = t[now << 1 | 1].all1 = 1;
		t[now << 1].num1 = mid - l + 1;
		t[now << 1 | 1].num1 = t[now].num1 - t[now << 1].num1;
		t[now << 1].all0 = t[now << 1 | 1].all0 = 0;
		t[now << 1].num0 = t[now << 1 | 1].num0 = 0;
	}
}

void build(int now, int l, int r) {
	if (l == r) {
		t[now].num0 = t[now].num1 = 0;
		t[now].all0 = t[now].all1 = 0;
		if (s[l]) t[now].num1++, t[now].all1 = 1;
			else t[now].num0++, t[now].all0 = 1;
		return ;
	}
	int mid = (l + r) >> 1;
	build(now << 1, l, mid);
	build(now << 1 | 1, mid + 1, r);
	up(now);
}

int get0(int now, int l, int r, int L, int R) {//找到这个区间 0/1 的个数
	if (L <= l && r <= R) {
		return t[now].num0;
	}
	down(now, l, r);
	int mid = (l + r) >> 1, re = 0;
	if (L <= mid) re += get0(now << 1, l, mid, L, R);
	if (mid < R) re += get0(now << 1 | 1, mid + 1, r, L, R);
	up(now);
	return re;
}

int get1(int now, int l, int r, int L, int R) {
	if (L <= l && r <= R) {
		return t[now].num1;
	}
	down(now, l, r);
	int mid = (l + r) >> 1, re = 0;
	if (L <= mid) re += get1(now << 1, l, mid, L, R);
	if (mid < R) re += get1(now << 1 | 1, mid + 1, r, L, R);
	up(now);
	return re;
}

void change0(int now, int l, int r, int L, int R) {//把这个区间全部改成 0/1
	if (L <= l && r <= R) {
		t[now].all0 = 1; t[now].num0 = r - l + 1;
		t[now].all1 = 0; t[now].num1 = 0;
		return ;
	}
	down(now, l, r);
	int mid = (l + r) >> 1;
	if (L <= mid) change0(now << 1, l, mid, L, R);
	if (mid < R) change0(now << 1 | 1, mid + 1, r, L, R);
	up(now);
}

void change1(int now, int l, int r, int L, int R) {
	if (L <= l && r <= R) {
		t[now].all1 = 1; t[now].num1 = r - l + 1;
		t[now].all0 = 0; t[now].num0 = 0;
		return ;
	}
	down(now, l, r);
	int mid = (l + r) >> 1;
	if (L <= mid) change1(now << 1, l, mid, L, R);
	if (mid < R) change1(now << 1 | 1, mid + 1, r, L, R);
	up(now);
}

bool find(int now, int l, int r, int pl) {//判断一个位置是 0/1
	if (l == r) {
		return t[now].num1;
	}
	down(now, l, r);
	int mid = (l + r) >> 1;
	if (pl <= mid) return find(now << 1, l, mid, pl);
		else return find(now << 1 | 1, mid + 1, r, pl);
}

bool ck(int mid) {//二分中的判断
	for (int i = 1; i <= n; i++)
		if (a[i] >= mid) s[i] = 1;
			else s[i] = 0;
	build(1, 1, n);
	
	for (int i = 1; i <= m; i++) {
		nm0 = get0(1, 1, n, q[i].l, q[i].r);
		nm1 = get1(1, 1, n, q[i].l, q[i].r);
		if (q[i].op == 0) {
			if (nm0) change0(1, 1, n, q[i].l, q[i].l + nm0 - 1);
			if (nm1) change1(1, 1, n, q[i].l + nm0, q[i].r);
		}
		else {
			if (nm1) change1(1, 1, n, q[i].l, q[i].l + nm1 - 1);
			if (nm0) change0(1, 1, n, q[i].l + nm1, q[i].r);
		}
	}
	
	return find(1, 1, n, pl);
}

int main() {
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
	for (int i = 1; i <= m; i++) {
		scanf("%d %d %d", &q[i].op, &q[i].l, &q[i].r);
	}
	scanf("%d", &pl);
	
	l = 1; r = n;
	while (l <= r) {//二分
		mid = (l + r) >> 1;
		if (ck(mid)) ans = mid, l = mid + 1;
			else r = mid - 1;
	}
	
	printf("%d", ans);
	
	return 0;
}

标签：4605,return,num1,jzoj,int,luogu,mid,num0,now
来源： https://blog.csdn.net/weixin_43346722/article/details/119193840