Codeforces Round #418 (Div. 2) C. An impassioned circulation of affection(尺取法)

//https://codeforces.com/contest/814/problem/B
//C. An impassioned circulation of affection
#include
using namespace std;
typedef long long ll;
const int maxn = 2e5 + 100;
vector<int>v[30];//存每个字符的位置
char s[maxn];
int main(){
	int n;
	cin >> n;
	scanf("%s", s + 1);
	for (int i = 1; i <= n; i++){
		int x = s[i] - 'a';
		v[x].push_back(i);
	}
	int q;
	cin >> q;
	for (int i = 1; i <= q; i++){
		int x;	char p;
		scanf("%d %c", &x, &p); 
		int ans = x;
		int k = p - 'a';
		if (v[k].size() == 0){
			printf("%d\n", x);
			continue;
		}
		if (n - v[k].size() <= x){
			printf("%d\n", n);
			continue;
		}
		int begin = 0 , end = 0;//起始 终点
		int cnt = 0;//记录走过多少边界 每次计算sum时需要减去cnt
		int sum = 0;//中间需要被修改的个数
		while (1){
			while (end < v[k].size()-1 && sum <= x)
				sum = v[k][++end] - v[k][begin] - 1 - cnt, cnt++;

			if (sum < x){
				ans = max(ans, x + cnt + 1);
				break;
			}
			if (sum >= x){//超过范围
				int h = sum - x;
				int num = 0;
				//判h==0因为可能右端点也包含
				ans = max(ans, v[k][end] - v[k][begin] - h+(h==0));
			}
			sum -= v[k][begin + 1] - v[k][begin]-1;
			begin++;
			if (begin >= v[k].size())
				break;
			cnt--;
		}
		printf("%d\n", ans);
	}
	return 0;
}