Mocha and Red and Blue

一、题目

题面翻译

给定长为 $n$ 的仅由 $\text{R}$、$\text{B}$、$\text{?}$ 组成的字符串 $S$，请你在 $\text{?}$ 处填入 $\text{R}$ 或 $\text{B}$，使得相邻位置字符相同的数量最少。

译者 @ajthreac

题目描述

As their story unravels, a timeless tale is told once again…

Shirahime, a friend of Mocha’s, is keen on playing the music game Arcaea and sharing Mocha interesting puzzles to solve. This day, Shirahime comes up with a new simple puzzle and wants Mocha to solve them. However, these puzzles are too easy for Mocha to solve, so she wants you to solve them and tell her the answers. The puzzles are described as follow.

There are $ n $ squares arranged in a row, and each of them can be painted either red or blue.

Among these squares, some of them have been painted already, and the others are blank. You can decide which color to paint on each blank square.

Some pairs of adjacent squares may have the same color, which is imperfect. We define the imperfectness as the number of pairs of adjacent squares that share the same color.

For example, the imperfectness of “BRRRBBR” is $ 3 $ , with “BB” occurred once and “RR” occurred twice.

Your goal is to minimize the imperfectness and print out the colors of the squares after painting.

输入格式

Each test contains multiple test cases.

The first line contains a single integer $ t $ ( $ 1 \le t \le 100 $ ) — the number of test cases. Each test case consists of two lines.

The first line of each test case contains an integer $ n $ ( $ 1\leq n\leq 100 $ ) — the length of the squares row.

The second line of each test case contains a string $ s $ with length $ n $ , containing characters ‘B’, ‘R’ and ‘?’. Here ‘B’ stands for a blue square, ‘R’ for a red square, and ‘?’ for a blank square.

输出格式

For each test case, print a line with a string only containing ‘B’ and ‘R’, the colors of the squares after painting, which imperfectness is minimized. If there are multiple solutions, print any of them.

样例 #1

样例输入 #1

5
7
?R???BR
7
???R???
1
?
1
B
10
?R??RB??B?

样例输出 #1

BRRBRBR
BRBRBRB
B
B
BRRBRBBRBR

提示

In the first test case, if the squares are painted “BRRBRBR”, the imperfectness is $ 1 $ (since squares $ 2 $ and $ 3 $ have the same color), which is the minimum possible imperfectness.

二、分析

找到第一个不是’?'的字符，向两边修改；

#include
#include
#include
using namespace std;
int main() {
	int T;
	cin >> T;
	while(T--) {
		int n;
		cin >> n;
		string s;
		cin >> s;
		if(count(s.begin(),s.end(),'B')== 0 && count(s.begin(),s.end(),'R')==0){ 
			for(int i = 0; i < s.size(); i++) {
				if(i % 2 == 0) s[i] = 'B';
				else s[i] = 'R';
			}
			cout<<s<<endl;continue;
		}
        for(int j=0;j<s.size();j++)
        {
            if(s[j]!='?'){
                for(int k=j-1;k>=0;k--)
                {
                    if(s[k]!='?') break;
                    if(s[k+1]=='B') s[k]='R';
                    else s[k]='B';
                }
                for(int k=j+1;k<s.size();k++)
                {
                    if(s[k]!='?') break;
                    if(s[k-1]=='B') s[k]='R';
                    else s[k]='B';
                }
            }
        }
        cout<<s<<endl;
    }
}

例题2

假如是计算相邻但不相等的对数，可以使用dp

样例输入 #1

5
7
?R???BR
7
???R???
1
?
1
B
10
?R??RB??B?

样例输出 #1

5
6
0
0
7

#include
#include
using namespace std;
const int N=200;
int dp[N][2]; //dp[i][0]表示第个位置选B的ans最大数量，dp[i][1]表示第i个位置选R的ans最大数量
int main()
{
    int T;cin>>T;
    while(T--)
    {
        memset(dp,0,sizeof(dp));
        int n;string s;
        cin>>n>>s;
        s=" "+s;
        for(int i=2;i<=n;i++)
        {
            if(s[i]=='?'||s[i]=='B')
            {
                dp[i][0]=max(dp[i-1][1]+1,dp[i-1][0]);
            }else{
                dp[i][0]=0;
            }
            if(s[i]=='?'||s[i]=='R')
            {
                dp[i][1]=max(dp[i-1][1]+1,dp[i-1][0]);
            }else{
                dp[i][1]=0;
            }
        }
        for(int i=1;i<=n;i++)
        cout<<dp[i][1]<<" ";
        cout<<endl;
        cout<<max(dp[n][0],dp[n][1])<<endl;
    }
}