C. Replacement（ Codeforces Round #316 (Div. 2) 模拟）-优快云博客

本文介绍了一个字符串处理问题，目标是在给定的字符串中通过特定的操作去除所有连续出现的两个点字符，同时处理多个查询更新字符串内容后的操作次数。

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Daniel has a string s, consisting of lowercase English letters and period signs (characters '.'). Let's define the operation of replacementas 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.")

题意：给一字符串，遇到两个连续的点【..】计数器+1，并变成一个点，把这样的操作叫做f(x)

m次查询每次给 X，C ---->把源字符串的x位置替换为字符c并计算一下当前字符串要进行多少次f(x)操作才能完全没有连续的2个点【.】

思路：先将初始的字符串需要多少步操作会变成要求的情况记录下来，记做sum，后面如果出现字母变成【.】，或者【.】变成字母，只需要判断一下这个位置的左右方向是不是【.】就行了。

复杂度O(n+4*m)

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

int n,m;
char str[300010];

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        scanf("%s",str);
        int sum = 0;
        for(int i=0;i<n-1;i++)
        {
            if(str[i] == '.' && str[i+1] == '.')
            {
                sum++;
            }
        }
        int x;
        char pp;
        while(m--)
        {
            scanf("%d%*c%c",&x,&pp);
            if((str[x-1]!='.' && pp!='.') || (str[x-1] == '.' && pp == '.'))
            {
                printf("%d\n",sum);
                continue;
            }
            str[x-1] = pp;
            if(pp != '.' && str[x-2] == '.')
            {
                sum--;
            }
            if(pp!='.' && str[x] == '.')
            {
                sum--;
            }
            if(pp == '.' && str[x-2] == '.')
            {
                sum++;
            }
            if(pp == '.' && str[x] == '.')
            {
                sum++;
            }
            printf("%d\n",sum);
        }
    }
    return 0;
}