UVA-679車的擺放(DFS)

                     ## 這是一道簡單的深度優先搜索問題##

附上題目鏈接：https://vjudge.net/problem/UVA-639
這道題跟八皇后問題類似，思路大致爲：把棋盤的每一格視作一個狀態，符合要求就擺放，不符合就進入下一狀態，到達最後一格就開始回溯，直至走完每一個狀態，由於是每一次都走到最後一格，符合深度優先搜索的思路。
附上AC代碼：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
int vis[4][4];//保存狀態，1爲放置，0爲不放
char bor[4][4];//保存棋盤狀態
int n, maxn;

/*判斷該格是否能放置車*/
bool judge(int x, int y){
    int flag = 1;
    if (bor[x][y] == 'X') return false;
    for (int i = x+1; i < n; i++){
        if (bor[i][y] == 'X') break;
        if (vis[i][y] == 1) flag = 0;
    }
    for (int i = x-1; i >= 0; i--){
        if (bor[i][y] == 'X') break;
        if (vis[i][y] == 1) flag = 0;
    }
    for (int i = y+1; i < n; i++){
        if (bor[x][i] == 'X') break;
        if (vis[x][i] == 1) flag = 0;
    }
    for (int i = y-1; i >= 0; i--){
        if (bor[x][i] == 'X') break;
        if (vis[x][i] == 1) flag = 0;
    }
    return flag;
}

void dfs(int cur, int cnt){
    if (cur == n*n){
        maxn = maxn>cnt?maxn:cnt;
        return;
    }
    if (judge(cur/n, cur%n)){
        vis[cur/n][cur%n] = 1;
        dfs(cur+1, cnt+1);
        vis[cur/n][cur%n] = 0;
        dfs(cur+1, cnt);
    }
    else dfs(cur+1, cnt);
}

int main()
{
    while(cin>>n&&n){
        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++)
                cin>>bor[i][j];
        for (int i = 0; i < n*n; i++){
            dfs(i, 0);
            memset(vis, 0, sizeof(vis));
        }
        cout<<maxn<<endl;
        maxn = 0;
        memset(bor, 0, sizeof(bor));
    }

    return 0;
}