C. Replacement
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Daniel has a strings, consisting of lowercase English letters and period signs (characters ‘.’). Let’s define the operation ofreplacementas the following sequence of steps: find a substring ".." (two consecutive periods) in strings, 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 stringscontains no two consecutive periods, then nothing happens.
Let’s definef(s)as the minimum number of operations ofreplacementto perform, so that the string does not have any two consecutive periods left.
You need to processmqueries, thei-th results in that the character at positionxi(1≤xi≤n) of stringsis assigned valueci. After each operation you have to calculate and output the value off(s).
Help Daniel to process all queries.
Input
The first line contains two integersnandm(1≤n,m≤300000) the length of the string and the number of queries.
The second line contains strings, consisting ofnlowercase English letters and period signs.
The followingmlines contain the descriptions of queries. Thei-th line contains integerxiandci(1≤xi≤n,ci— a lowercas English letter or a period sign), describing the query of assigning symbolcito positionxi.
Output
Printmnumbers, one per line, thei-th of these numbers must be equal to the value off(s)after performing thei-th assignment.
Sample test(s)
input
10 3.b..bz….1 h3 c9 f
output
431
input
4 4.cc.2 .3 .2 a1 a
output
1311
Note
Note to the first sample test (replaced periods are enclosed in square brackets).
The original string is ".b..bz….".
after the first query= 4("hb[..]bz…."→"hb.bz[..].."→"hb.bz[..]."→"hb.bz[..]"→"hb.bz.")after the second query= 3("hbс.bz[..].."→"hbс.bz[..]."→"hbс.bz[..]"→"hbс.bz.")after the third query= 1("hbс.bz[..]f."→"hbс.bz.f.")
Note to the second sample test.
The original string is ".cc.".
after the first query:= 1("[..]c."→".c.")after the second query:= 3("[..].."→"[..]."→"[..]"→".")after the third query:= 1(".a[..]"→".a.")after the fourth query:= 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)
未经一番寒彻骨，焉得梅花扑鼻香
