zoukankan      html  css  js  c++  java
  • Codeforces 570C. Replacement

    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 ≤ nci — 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;
    }
    View Code
  • 相关阅读:
    linux 下java jar包的方法
    (转)浅谈Java的输入输出流
    把java文件打包成jar文件
    C#ListView控件中列添加控件ComboBox,控件TextBox,添加时间选择列DateTimePicker
    <LabelId>k__BackingField反编译错误修改
    oracleI基础入门(6)sql语句And or Crazy
    oracleI基础入门(6)sql语句distinct Crazy
    oracleI基础入门(6)sql语句Order By Crazy
    oracleI基础入门(6)sql语句Like Crazy
    oracleI基础入门(6)sql语句count Crazy
  • 原文地址:https://www.cnblogs.com/ToTOrz/p/7007613.html
Copyright © 2011-2022 走看看