PAT--1057 Stack

题目链接：https://pintia.cn/problem-sets/994805342720868352/problems/994805417945710592

题目大意：模仿栈，只不过在原有的Push，Pop操作上，又加了一个PeekMedian操作，即当栈内的数是偶数时，输出第（n / 2）小的数，当栈内的数是奇数时，输出第（n + 1）/ 2小的数。

分析：树状数组+二分，或权值线段树（坤坤一语道破天机）。

树状数组+二分

#include <bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
const int N = 1e5 + 5;
int n, c[N];
char s[15];
stack<int> q;
int lowbit(int x) {
	return x & (-x);
}
void update(int x, int w) {
	while(x < N) {
		c[x] += w;
		x += lowbit(x);
	}
}
int ask(int x) {
	int sum = 0;
	while(x) {
		sum += c[x];
		x -= lowbit(x);
	}
	return sum;
}
int find(int x) {
	int l = 0, r = 1e5, ans;
	while(l <= r) {
		int mid = (l + r) >> 1;
		if(ask(mid) < x) {
			l = mid + 1;
		} else {
			r = mid - 1;
			ans = mid;
		}
	}
	return ans;
}
int main() {
	scanf("%d", &n);
	memset(c, 0, sizeof c);
	while(n--) {
		scanf("%s", s);
		if(s[1] == 'u') {
			int x;
			scanf("%d", &x);
			q.push(x);
			update(x, 1);
		} else if(s[1] == 'o') {
			if(q.size() == 0) {
				printf("Invalid\n");
			} else {
				int x = q.top();
				q.pop();
				printf("%d\n", x);
				update(x, -1);
			}
		} else {
			if(q.size() == 0) {
				printf("Invalid\n");
			} else {
				int x = (q.size() + 1) / 2;
				printf("%d\n", find(x));
			}
		}
	}
	return 0;
}

权值线段树

#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int n;
string s;
stack<int> q;
struct node {
	int l, r, sum;
}tr[N<<2];
void pushup(int m) {
	tr[m].sum = tr[m<<1].sum + tr[m<<1|1].sum;
}
void build(int m, int l, int r) {
	tr[m].l = l;
	tr[m].r = r;
	tr[m].sum = 0;
	if(l == r) return ;
	int mid = (l + r) >> 1;
	build(m<<1, l, mid);
	build(m<<1|1, mid + 1, r);
}
void update(int m, int id, int w) {
	if(tr[m].l == id && tr[m].r == id) {
		tr[m].sum += w;
		return ;
	}
	int mid = (tr[m].l + tr[m].r) >> 1;
	if(id <= mid) update(m<<1, id, w);
	else update(m<<1|1, id, w);
	pushup(m);
}
int ask(int m, int num) {
	if(tr[m].l == tr[m].r) return tr[m].l;
	if(tr[m<<1].sum >= num) return ask(m<<1, num);
	else return ask(m<<1|1, num - tr[m<<1].sum);
}
int main() {
	build(1, 1, N);
	scanf("%d", &n);
	while(n--) {
		cin >> s;
		if(s == "Push") {
			int x;
			scanf("%d", &x);
			q.push(x);
			update(1, x, 1);
		} else if(s == "Pop") {
			if(q.size() == 0) printf("Invalid\n");
			else {
				int x = q.top();
				q.pop();
				printf("%d\n", x);
				update(1, x, -1);
			}
		} else {
			if(q.size() == 0) printf("Invalid\n");
			else {
				int x = q.size();
				x = (x + 1) / 2;
				printf("%d\n", ask(1, x));
			}
		}
	}
	return 0;
}