D - Face Produces Unhappiness

atcoder-ABC140

The question ：

1. If there is s[i-1]=s[i],happy The number of people +1
2. What we can do is , Select an interval to reverse the whole character （L change R）
3. At most k operations （ Note that not necessarily K operations ）

Ideas ：

1. We first record the original string happy The number of people ( as long as s[i-1]==s[i], namely LL Type or RR type , Because we only discuss the properties of adjacent two at a time )
2. Then the rest is in the string , only LR The type and RL type
   eg:LRLRL perhaps RLRL（ At least three characters or more are such LRL or RLR type ）, We will find that whether it is flipping L perhaps R, Can get three consecutive equal characters , bring happy The number of +2
3. So we added happy The number is 2*k, however ！ The most in a string happy The number of people is n-1 personal , What is given k For the most operations , It may be much larger than the maximum operation we can actually carry out ！
   So we need to take happy=min(happy+2*k,n-1);

#include<iostream>
#include<cstring>
using namespace std;
int main()
{
    
	int n,k;
	cin>>n>>k;
	string s;
	cin>>s;
	int happy=0;
	for(int i=1;i<n;i++)
	if(s[i]==s[i-1]) happy++;
	cout<<min(happy+2*k,n-1)<<endl;
	
	return 0;
}