给个样例
8 2333
D 2
D 4
D 2
R
Q 1 //这个查询应该是在hdu是3，在POJ上是3或1都能过。我亲测的。两个OJ的测试数据不同。同一个村庄可能摧毁几次，修复一次，是否可以把前些次D操作全修好呢？
下面是两个OJ都可以AC的代码

#include <iostream> 
#include <cstring>
using namespace std;
const int N = 50007;
int bit[N], n, s[N];
bool vis[N];
inline int lowbit(int x) {
	return x & -x;
}
void add(int pos, int val) {
	for (int i = pos; i <= n; i += lowbit(i))
		bit[i] += val;
}
int sum(int pos) {
	int res = 0;
	for (int i = pos; i > 0; i -= lowbit(i))
		res += bit[i];
	return res;
}
int main()
{
	ios::sync_with_stdio(false);
	int q;
	while (cin >> n >> q) {
		memset(bit, 0, sizeof bit);
		memset(vis, 1, sizeof vis);
		int cur = 0;
		for (int i = 1; i <= n; ++i) //这一步有点耗时，若能把0看做连通，1看做非连通，初始全为0可以用memset
			add(i, 1);
		char opt; int x = 0;
		while (q--) {
			cin >> opt; if (opt != 'R') cin >> x;
			if (opt == 'D') {
				s[cur++] = x;
				if (vis[x]) {
					vis[x] = false;					
					add(x, -1);
				}
			}
			else if (opt == 'R') {
				if (cur > 0) {
					x = s[--cur];
					add(x, 1);
					vis[x] = true;
				}
			}
			else if (opt == 'Q') {
				if (!vis[x]) {
					cout << 0 << endl;
					continue;
				}
				int tmp = sum(x), a = 1, b = n;
				int l = 0, r = x; //左侧连通数
				while (l <= r) {
					int m = l + r >> 1;
					if (tmp - sum(m) == x - m) {
						a = m + 1;
						r = m - 1;
					}
					else l = m + 1;
				}
				l = x, r = n;
				while (l <= r) {
					int m = l + r >> 1;
					if (sum(m) - tmp == m - x) {
						b = m;
						l = m + 1;
					}
					else r = m - 1;
				}
				cout << b - a + 1 << endl;
			}
		}
	}

	return 0;
}