暴力枚举多少行白、多少行蓝

#include <bits/stdc++.h>
using namespace std;
int n, m;
char g[55][55];
int main()
{
    // 输入
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            cin >> g[i][j];
    // 枚举所有情况
    int ans = n * m;
    for (int a = 1; a <= n - 2; a++)
        for (int b = 1; b <= n - a - 1; b++)
        {
            int c = n - a - b;
            // a行白色、b行蓝色、c行红色，要涂几次
            int now = 0;
            // 1~a 行有几个不是白色 a+1~a+b 行有几个不是蓝色，a+b+1~n 行有几个不是红色
            for (int i = 1; i <= a; i++)
                for (int j = 1; j <= m; j++)
                    if (g[i][j] != 'W')
                        now++;
            for (int i = a + 1; i <= a + b; i++)
                for (int j = 1; j <= m; j++)
                    if (g[i][j] != 'B')
                        now++;
            for (int i = a + b + 1; i <= n; i++)
                for (int j = 1; j <= m; j++)
                    if (g[i][j] != 'R')
                        now++;
            ans = min(ans, now);
        }
    cout << ans;
    return 0;
}

提前预处理每行每种颜色的数量

#include <bits/stdc++.h>
using namespace std;
int n, m;
char g[55][55];
int cnt[55][3]; // cnt[i][0/1/2] 第 i 行白蓝红的数量
int main()
{
    // 输入
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
        {
            cin >> g[i][j];
            if (g[i][j] == 'W')
                cnt[i][0]++;
            else if (g[i][j] == 'B')
                cnt[i][1]++;
            else
                cnt[i][2]++;
        }
    // 枚举所有情况
    int ans = n * m;
    for (int a = 1; a <= n - 2; a++)
        for (int b = 1; b <= n - a - 1; b++)
        {
            int c = n - a - b;
            // a行白色、b行蓝色、c行红色，要涂几次
            int now = 0;
            // 1~a 行有几个不是白色 a+1~a+b 行有几个不是蓝色，a+b+1~n 行有几个不是红色
            for (int i = 1; i <= a; i++)
                now += cnt[i][1] + cnt[i][2];
            for (int i = a + 1; i <= a + b; i++)
                now += cnt[i][0] + cnt[i][2];
            for (int i = a + b + 1; i <= n; i++)
                now += cnt[i][0] + cnt[i][1];
            ans = min(ans, now);
        }
    cout << ans;
    return 0;
}

前缀和优化区间和的计算

略