ZOJ 1002

题目很简单，就是类似八皇后的dfs问题，但是有着限制条件。

所以要在dfs过程中check 是否符合条件

代码如下：

#include <iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<stack>
#include<cstring>
using namespace std;
int MAX;
int a;
int now;
char str[4][4];
//判断能不能放
int canPut(int i, int j)
{
	int ok = 1;
	if (str[i][j] == '1'||str [i][j]=='X') ok = 0;
	for (int k = i; k >= 0; k--)
	{
		if (str[k][j] == 'X')  break;
		if (str[k][j] == '1')  ok = 0;
	}

	for (int k = i; k < a; k++)
	{
		if (str[k][j] == 'X')  break;
		if (str[k][j] == '1')  ok = 0;
	}

	for (int k = j; k >= 0; k--)
	{
		if (str[i][k] == 'X')  break;
		if (str[i][k] == '1')  ok = 0;
	}

	for (int k = j; k < a; k++)
	{
		if (str[i][k] == 'X')  break;
		if (str[i][k] == '1')  ok = 0;
	}
	return ok;
}

void dfs(int now,int next) {
	
	
	for (int x=next; x < a; x++) {
		for (int y = 0;y < a;y++) {
			if (canPut(x,y)) {
				str[x][y] ='1';//放堡垒
				now++;
				if (MAX < now)//更新
					MAX = now;

				dfs(now,x);//放下一座堡垒

				str[x][y]='.';//拿开堡垒
				now--;
			}
		}
	}
	return;
}

int main() {
	while (cin >> a) {
		MAX= 0;
		if (a == 0)//end of input
			break;
		memset(str, 0, sizeof(str));
		for (int i = 0; i <a; i++) {
			cin >> str[i];
		}
		dfs(0,0);
		cout << MAX<<endl;
	}
	return 0;
}