Codeforces Round #316 (Div. 2)570C Replacement(脑洞)

C. Replacement

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Daniel has a string s, consisting of lowercase English letters and period signs (characters '.'). Let's define the operation of replacement as the following sequence of steps: find a substring ".." (two consecutive periods) in string s, of all occurrences of the substring let's choose the first one, and replace this substring with string ".". In other words, during the replacement operation, the first two consecutive periods are replaced by one. If string s contains no two consecutive periods, then nothing happens.

Let's define f(s) as the minimum number of operations of replacement to perform, so that the string does not have any two consecutive periods left.

You need to process m queries, the i-th results in that the character at position xi (1 ≤ xi ≤ n) of string s is assigned value ci. After each operation you have to calculate and output the value of f(s).

Help Daniel to process all queries.

Input

The first line contains two integers n and m (1 ≤ n, m ≤ 300 000) the length of the string and the number of queries.

The second line contains string s, consisting of n lowercase English letters and period signs.

The following m lines contain the descriptions of queries. The i-th line contains integer xi and ci (1 ≤ xi ≤ n, ci — a lowercas English letter or a period sign), describing the query of assigning symbol ci to position xi.

Output

Print m numbers, one per line, the i-th of these numbers must be equal to the value of f(s) after performing the i-th assignment.

Sample test(s)

input

10 3
.b..bz....
1 h
3 c
9 f

output

4
3
1

input

4 4
.cc.
2 .
3 .
2 a
1 a

output

1
3
1
1

Note

Note to the first sample test (replaced periods are enclosed in square brackets).

The original string is ".b..bz....".

• after the first query f(hb..bz....) = 4    ("hb[..]bz...."  →  "hb.bz[..].."  →  "hb.bz[..]."  →  "hb.bz[..]"  → "hb.bz.")
• after the second query f(hbс.bz....) = 3    ("hbс.bz[..].."  →  "hbс.bz[..]."  →  "hbс.bz[..]"  →  "hbс.bz.")
• after the third query f(hbс.bz..f.) = 1    ("hbс.bz[..]f."  →  "hbс.bz.f.")

Note to the second sample test.

The original string is ".cc.".

• after the first query: f(..c.) = 1    ("[..]c."  →  ".c.")
• after the second query: f(....) = 3    ("[..].."  →  "[..]."  →  "[..]"  →  ".")
• after the third query: f(.a..) = 1    (".a[..]"  →  ".a.")
• after the fourth query: f(aa..) = 1    ("aa[..]"  →  "aa.")

给一个长度为n的字符串，m次动作，问你每次变化后都多少个".."

主要思考的是如何避免TLE，可以计算出第一次的ans，而后每次改变借用前一次的ans即可，若每次改变为由字母变为 '.' 且前或后有

 '.' ans就加1，由 '.' 变成字母则相反。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int MAXN = 300005;
char s[MAXN], str[10];
int n, m, x, ans;
bool flag;
int main(int argc, char const *argv[])
{
	scanf("%d%d", &n, &m);
	scanf("%s", s + 1);
	scanf("%d %s", &x, str);
	s[x] = str[0];
	for(int i = 1; i <= n; ++i) {
		int cur = 0;
		flag = false;
		while(s[i] == '.') {
			i++;
			cur++;
			flag = true;
		}
		if(flag) ans += cur - 1;
	}
	printf("%d\n", ans);
	m--;
	while(m--) {
		scanf("%d %s", &x, str);
		char c = s[x];
		s[x] = str[0];
		int cur = 0;
		if(str[0] == '.' && c != '.') {
			if(s[x - 1] == '.') ans++;
			if(s[x + 1] == '.') ans++;
 		}
 		else if(str[0] != '.' && c == '.') {
 			if(s[x - 1] == '.') ans--;
 			if(s[x + 1] == '.') ans--;
 		}
		printf("%d\n", ans);
	}
	return 0;
}