【贪心思想】Blocks

题目

 题目来源：（校选拔）

NTU_ACM-2021新生预选赛2 - Virtual Judgehttps://vjudge.net/contest/462138#problem/ZProblem - B - Codeforceshttps://codeforces.com/contest/1271/problem/B

解析

感谢这篇文章的思路与代码提供

B. Blocks-------思维+贪心(稍思维)_AKone123456的博客-CSDN博客There are n blocks arranged in a row and numbered from left to right, starting from one. Each block is either black or white.You may perform the following operation zero or more times: choose two adj...https://blog.csdn.net/qq_43690454/article/details/104093824?ops_request_misc=&request_id=&biz_id=102&utm_term=There%20are%20nn%20blocks%20arranged%20i&utm_medium=distribute.pc_search_result.none-task-blog-2~all~sobaiduweb~default-2-104093824.first_rank_v2_pc_rank_v29&spm=1018.2226.3001.4187我们可以采取贪心的思想，首先认为方块可以全部变成白色，这样在第一次遍历中将遇到黑色的方块就翻转一次，到最后后变成n-1个W+B/W，如果最后一个方块黑色，我们再从头进行一次翻转，认为全部可以翻成黑色，此时如果n-1为偶数的话则可以反转，n-1为奇数的话则输出-1

ac源代码

const int N = 1e5 + 1000;
int n;
int path[N];
vector<int> v;
char s[N];
int tot;
int main()
{
    cin >> n;
    cin >> (s + 1);
    //假设全白
    for (int i = 1; i < n; i++)
    {
        if (s[i] == 'B')
        {
            s[i] = 'W';
            s[i + 1] = s[i + 1] == 'W' ? 'B' : 'W';
            v.push_back(i);
        }
    }
    if (s[n] == 'W')
    {
        cout << v.size() << endl;
        for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
        cout << endl;
        return 0;
    }
    //假设全黑 
    for (int i = 1; i < n; i++)
    {
        if (s[i] == 'W')
        {
            s[i] = 'B';
            s[i + 1] = s[i + 1] == 'W' ? 'B' : 'W';
            v.push_back(i);
        }
    }
    if (s[n] == 'B')
    {
        cout << v.size() << endl;
        for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
        cout << endl;
        return 0;
    }
    cout << -1 << endl;
    return 0;
}