UVa11214 Guarding the Chessboard (IDA*)

链接：http://vjudge.net/problem/23958

分析：控制最大递归深度为maxd，也就是枚举个数maxd个皇后守卫的时候，dfs以行枚举放皇后守卫的位置，确保放皇后守卫的位置的行，列，主和副对角线上没有其它守卫，否则就不是最优的了，每次递归从上次放的行的下一行开始放，放满maxd个后进行判断是否所有标记格子都被攻击到。是则直接返回true输出最少守卫，否则返回递归上一层继续枚举，vis要复位。

 1 #include <cstdio>
 2 #include <cstring>
 3
 4 const int maxn = 11;
 5
 6 int n, m, maxd;
 7 char G[maxn][maxn];
 8 bool vis[4][maxn << 1];
 9
10 bool dfs(int cur, int d) {
11     if (d == maxd) {
12         for (int i = 1; i <= n; i++)
13             for (int j = 1; j <= m; j++)
14                 if (G[i][j] == ‘X‘ && !vis[0][i] && !vis[1][j] && !vis[2][j - i + n] && !vis[3][i + j])
15                     return false;
16         return true;
17     }
18     for (int i = cur; i <= n; i++)
19         for (int j = 1; j <= m; j++) {
20             if (!vis[0][i] || !vis[1][j] || !vis[2][j - i + n] || !vis[3][i + j]) {
21                 bool br = vis[0][i], bc = vis[1][j], bm = vis[2][j - i + n], bd = vis[3][i + j];
22                 vis[0][i] = vis[1][j] = vis[2][j - i + n] = vis[3][i + j] = true;
23                 if (dfs(cur + 1, d + 1)) return true;
24                 vis[0][i] = br, vis[1][j] = bc, vis[2][j - i + n] = bm, vis[3][i + j] = bd;
25             }
26         }
27     return false;
28 }
29
30 int main() {
31     int kase = 0;
32     while (scanf("%d%d", &n, &m) == 2 && n) {
33         for (int i = 1; i <= n; i++) scanf("%s", G[i] + 1);
34         for (maxd = 0; ; maxd++) {
35             memset(vis, 0, sizeof(vis));
36             if (dfs(1, 0)) break;
37         }
38         printf("Case %d: %d\n", ++kase, maxd);
39     }
40     return 0;
41 }