hdoj 5319 Painter(模拟题)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5319

思路分析：假设颜色R表示为1，颜色B表示为2，颜色G表示为3，因为数据量较小，采用暴力解法即可，即每次扫描对角线，看每条对角线需要画多少笔，统计所有对角线的笔数和即可；

代码如下：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int MAX_N = 50 + 10;
int map[MAX_N][MAX_N];

int Solve(int n, int m)
{
    // m 表示行, n表示列数
    int count = 0;
    bool flag = false;
    for (int k = m - 1; k >= 1 - n; -- k)
    {
        flag = false;
        for (int j = 0; j < n; ++ j)
        {
            int i = j + k;
            if (0 <= i && i < m && 0 <= j && j < n)
            {
                if (map[i][j] & 1)
                {
                    map[i][j] = map[i][j] - 1;
                    if (!flag)
                    {
                        flag = true;
                        count++;
                    }
                }  else if ((map[i][j] & 1) == 0 && flag)
                    flag = false;
            }
        }
    }
    flag = false;
    int t = m + n;
    for (int k = 0; k < t; ++ k)
    {
        flag = false;
        for (int j = 0; j < n; ++ j)
        {
            int i = -j + k;
            if (0 <= i && i < m && 0 <= j && j < n)
            {
                if (map[i][j] & 2)
                {
                    map[i][j] = map[i][j] - 2;
                    if (!flag)
                    {
                        flag = true;
                        count++;
                    }
                } else if ((map[i][j] & 2) == 0 && flag)
                    flag = false;
            }
        }
    }
    return count;
}

int main()
{
    int case_times;
    char str[MAX_N];
    int n, m;

    scanf("%d", &case_times);
    while (case_times--)
    {
        scanf("%d", &n);
        memset(map, 0, sizeof(map));
        for (int i = 0; i < n; ++ i)
        {
            scanf("%s", str);
            int len = m = strlen(str);
            for (int j = 0; j < len; ++ j)
            {
                if (str[j] == ‘.‘)
                    map[i][j] = 0;
                if (str[j] == ‘R‘)
                    map[i][j] = 1;
                if (str[j] == ‘B‘)
                    map[i][j] = 2;
                if (str[j] == ‘G‘)
                    map[i][j] = 3;
            }
        }

        int ans = Solve(m, n);
        printf("%d\n", ans);

    }
    return 0;
}