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.
f(s)就是所有连续的‘.‘序列合起来的长度-序列的数量
每次处理合并时观察两边的序列是否能合起来
#include<iostream>
#include<cstring>
#include<cstdio>
#include <string>
#include <sstream>
#include <map>
#include <cmath>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
#include <set>
using namespace std;
typedef long long LL;
#define MOD 1000000007
string s;
int n,m;
int ok[300005];
int main()
{
freopen("test.in","r",stdin);
cin >> n >> m;
cin >> s;
LL group(0),num(0);
for (int i=0;i<n;i++){
if (s[i] == ‘.‘){
num ++;
if (i == 0 || s[i-1] != ‘.‘){
group ++;
}
ok[i+1] = 1; // ok[i] 表示该位置是否为i
}
}
for (int i=1;i<=n;i++){
cout << ok[i];
}cout << endl;
for (int i=1;i<=m;i++){
string ts;
int d;
cin >> d >> ts;
char c = ts[0];
int a = ok[d], b = (c == ‘.‘);
if (a != b){
if (a){
num --;
}
else {
num ++;
}
if (ok[d-1] && ok[d+1] && !b){
group ++;
}
if (ok[d-1] && ok[d+1] && b){
group --;
}
if (!ok[d-1] && !ok[d+1] && !b){
group --;
}
if (!ok[d-1] && !ok[d+1] && b){
group ++;
}
}
ok[d] = b;
cout << num - group << endl;
}
return 0;
}