例题1.14 填充正方形 UVa11520

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1.题目描述：点击打开链接

2.解题思路：本题要求字典序最小，由于n的范围比较小，直接尝试按照字典序填写每个格子即可。

3.代码：

#define _CRT_SECURE_NO_WARNINGS #include<iostream>#include<algorithm>#include<string>#include<sstream>#include<set>#include<vector>#include<stack>#include<map>#include<queue>#include<deque>#include<cstdlib>#include<cstdio>#include<cstring>#include<cmath>#include<ctime>#include<functional>using namespace std;#define N 10+5char g[N][N];int dx[] = { 1, -1, 0, 0 };int dy[] = { 0, 0, 1, -1 };int vis[26];int n;bool is_inside(int r, int c){if (r < 0 || r >= n || c < 0 || c >= n)return false;return true;}void dfs(int cur){if (cur == n*n){for (int i = 0; i < n;i++)for (int j = 0; j < n; j++)printf("%c%s", g[i][j], j == n - 1 ? "\n" : "");return;}int r = cur / n, c = cur%n;if (g[r][c] == '.'){memset(vis, 0, sizeof(vis));for (int d = 0; d < 4; d++){int nx = r + dx[d], ny = c + dy[d];if (is_inside(nx, ny) && isupper(g[nx][ny]))//找周围的字母vis[g[nx][ny] - 'A'] = 1;}for (int i = 0; i<26; i++)//找字典序最小的字母if (!vis[i]){ g[r][c] = i + 'A'; break; }}dfs(cur + 1);}int main(){//freopen("t.txt", "r", stdin);int T;cin >> T;for (int rnd = 1; rnd <= T; rnd++){scanf("%d", &n);for (int i = 0; i < n; i++)scanf("%s", g[i]);printf("Case %d:\n",rnd);dfs(0);}return 0;}

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