专题一搜索 G - Guarding the Chessboard

1. 题目
   Given an n ∗ m chessboard with some marked squares, your task isto place as few queens as possible to guard (attack or occupy) allmarked squares. Below is a solution to an 8 ∗ 8 board with everysquare marked. Note that queens can be placed on non-markedsquares.
   
   Input
   The input consists of at most 15 test cases. Each case begins witha line containing two integers n, m (1 < n, m < 10) the size ofthe chessboard. Next n lines each contain m characters, ‘X’ denotesmarked square, ‘.’ denotes unmarked squares. The last case isfollowed by a single zero, which should not be processed.
   Output
   For each test case, print the case number and the minimal number of queens needed.
   
   Sample Input
   8 8
   XXXXXXXX
   XXXXXXXX
   XXXXXXXX
   XXXXXXXX
   XXXXXXXX
   XXXXXXXX
   XXXXXXXX
   XXXXXXXX
   8 8
   X.......
   .X......
   ..X.....
   ...X....
   ....X...
   .....X..
   ......X.
   .......X
   0
   Sample Output
   Case 1: 5
   Case 2: 1
2. 思路
   八皇后问题的改编，不过看到有点低的通过数我意识到这玩意可能没那么简单
   最后用了个东西叫迭代加深搜索
   简单点说：八皇后问题一般都是dfs（算是dfs样板题），而为了在有限时间以内得到最优结果，每次深搜的时候限制搜索深度，依次尝试只放一个皇后两个皇后n个皇后可不可行，可行时停止加深输出结果。此外运行一次可以知道10*10放满的情况下最多5个皇后就能保卫棋盘，所以搜完4个皇后还没得到结果的话，答案肯定是5个皇后。
3. 代码
   

   #include<cstdio>
   #include<cstring>
   #include<cmath>
   #include<algorithm>
   #include<queue>
   using namespace std;
   
   int n,m;
   int vis[4][25];
   int dep;//当前深度
   char map[15][15];
   
   bool dfs(int num,int r)//个数，行数
   {
   	if(num>dep)
   	{
   		for(int i=1;i<=n;i++)
   		{
   			for(int j=1;j<=m;j++)
   			if(map[i][j]=='X'&&!vis[0][i]&&!vis[1][j]&&!vis[2][i+j]&&!vis[3][i-j+m])
   				return 0;
   		}
   		return 1;
   	}	
   	for(int i=r;i<=n;i++)//不一定每行都能放，比八皇后多一个循环
   	{
   		for(int j=1;j<=m;j++)
   		{
   			if(!vis[0][i]||!vis[1][j]||!vis[2][i+j]||!vis[3][i-j+m])
   			{
   				int v1=vis[0][i],v2=vis[1][j],v3=vis[2][i+j],v4=vis[3][i-j+m];
   				vis[0][i]=1, vis[1][j]=1, vis[2][i+j]=1, vis[3][i-j+m]=1;
   				if(dfs(num+1,i+1)) return 1;
   				vis[0][i]=v1,vis[1][j]=v2,vis[2][i+j]=v3,vis[3][i-j+m]=v4;
   			}
   		}
   	}
   	return 0;
   }
   
   int main()
   {
   	int q=1;
   	while(scanf("%d",&n)!=EOF&&n!=0)
   	{
   		scanf("%d",&m);
   		getchar();
   		for(int i=1;i<=n;i++){
   			for(int j=1;j<=m;j++){
   				scanf("%1c",&map[i][j]);
   			}
   			getchar();
   		}
   		for(dep=1;dep<5;dep++)//10*10放满只需要5个皇后，搜完4个还没结果直接出答案
   		{
   			memset(vis,0,sizeof(vis));
   			if(dfs(1,1))  break;
   		}
   		printf("Case %d: %d\n",q++,dep);
   	}
   	return 0;
   }