二分查找

先把障碍物坐标放入一维数组中，再每次查询时，二分查找左端和右端+1的位置，如果两个迭代器相同或查找失败就返回0，否则输出答案长度。

AC 代码

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define all(a) a.begin(), a.end()
#define fi first
#define se second
#define pb push_back
#define pp pop_back
#define pf push_front
#define lb lower_bound
#define ub upper_bound
#define UNIQUE(v) sort(all(v)); v.erase(unique(all(v)), v.end())
#define all(v) v.begin(), v.end()
#define sz(v) (int)v.size()
#define PQ priority_queue
constexpr ll inf = 1e9 + 7;

void solve() {
	int n, q;
	cin >> n >> q;
	string s;
	cin >> s;
	vector<int> vec;
	for (int i = 0; i < n; i++) {
		if (s[i] == '#') vec.pb(i+1);
	}
	while (q--) {
		int x, y;
		cin >> x >> y;
		int l = min(x, y);
		int r = max(x, y);	
		auto it1 =lb(all(vec), l);
		auto it2 = ub(all(vec), r);
		if (it1 == vec.end() || it1 == it2) {
			cout << 0 << "\n";
			continue;
		}
		it2--;
		cout << *it2 - *it1 + 1 << "\n";
	}
}

int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--) solve();
    return 0;
}