Codeforces Round #316 (Div. 2) C 模擬

連結：戳這裡

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.

Examples
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.")

題意：

長度為n的字串，連續兩個都為'.'的話算一個價值

給q個操作x,c，每次將位置x的字元換成c

問當前的字串價值

思路：

當前如果是小寫字母變成'.'則算增加的價值，同理'.'變成小寫字母算減少的價值

程式碼：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int n,m;
int a[300100];
string s;
int main(){
    scanf("%d%d",&n,&m);
    cin>>s;
    for(int i=0;i<s.size();i++){
        if(s[i]=='.') a[i+1]=1;
        else a[i+1]=0;
    }
    int ans=0;
    for(int i=1;i<=n;i++){
        while(a[i]==1 && a[i+1]==1 && i+1<=n ){
            i++;
            ans++;
        }
    }
    for(int i=1;i<=m;i++){
        int x;
        char c;
        scanf("%d %c",&x,&c);
        if(c=='.'){
            if(a[x]==1) {
                cout<<ans<<endl;
                continue;
            }
            a[x]=1;
            if(x==1 && a[2]==1) ans++;
            else if(x==n && a[n-1]==1) ans++;
            else {
                if(a[x-1]==1) ans++;
                if(a[x+1]==1) ans++;
            }
        } else {
            if(a[x]==0){
                cout<<ans<<endl;
                continue;
            }
            a[x]=0;
            if(x==1 && a[2]==1) ans--;
            else if(x==n && a[n-1]==1) ans--;
            else {
                if(a[x-1]==1) ans--;
                if(a[x+1]==1) ans--;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}