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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; }
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原文地址：http://www.cnblogs.com/tallisHe/p/4683643.html
