DFS+模拟 ZOJ 3861 Valid Pattern Lock

 

题目传送门

 1 /*  2  题意：手机划屏解锁，一笔连通所有数字,输出所有可能的路径；  3  DFS：全排列 + ok () 判断函数，去除一些不可能连通的点:)  4 */  5 #include <cstdio>  6 #include <iostream>  7 #include <algorithm>  8 #include <cstring>  9 #include <string>  10 #include <map>  11 #include <cmath>  12 #include <vector>  13 #include <set>  14 #include <queue>  15 using namespace std;  16  17 const int MAXN = 500000;  18 const int INF = 0x3f3f3f3f;  19 int a[11], ans[11];  20 int n, x;  21 int vis[11];  22 int res[MAXN][11];  23 int flag[10];  24  25 bool ok(void)  26 {  27 memset (flag, 0, sizeof (flag));  28  29 int i;  30 for (i=1; i<=n-1; i++) //小细节决定成败  31  {  32 flag[ans[i]] = 1;  33 if((ans[i] == 1 && ans[i+1] == 9 && flag[5] == 0)|| (ans[i] == 9 && ans[i+1] == 1 && flag[5] == 0) ||  34 (ans[i] == 1 && ans[i+1] == 3 && flag[2] == 0)|| (ans[i] == 3 && ans[i+1] == 1 && flag[2] == 0) ||  35 (ans[i] == 1 && ans[i+1] == 7 && flag[4] == 0)|| (ans[i] == 7 && ans[i+1] == 1 && flag[4] == 0) ||  36 (ans[i] == 2 && ans[i+1] == 8 && flag[5] == 0)|| (ans[i] == 8 && ans[i+1] == 2 && flag[5] == 0) ||  37 (ans[i] == 4 && ans[i+1] == 6 && flag[5] == 0)|| (ans[i] == 6 && ans[i+1] == 4 && flag[5] == 0) ||  38 (ans[i] == 3 && ans[i+1] == 9 && flag[6] == 0)|| (ans[i] == 9 && ans[i+1] == 3 && flag[6] == 0) ||  39 (ans[i] == 7 && ans[i+1] == 9 && flag[8] == 0)|| (ans[i] == 9 && ans[i+1] == 7 && flag[8] == 0) ||  40 (ans[i] == 3 && ans[i+1] == 7 && flag[5] == 0)|| (ans[i] == 7 && ans[i+1] == 3 && flag[5] == 0) )  41 return 0;  42  43  }  44 if(i == n)  45 return true;  46 }  47  48 void DFS(int cnt)  49 {  50 if (cnt == n + 1)  51  {  52 if (ok ())  53  {  54 x++;  55 for (int i=1; i<=n; i++)  56  {  57 res[x][i] = ans[i];  58  }  59  }  60  61 return ;  62  }  63  64 for (int i=1; i<=n; i++)  65  {  66 if (!vis[a[i]])  67  {  68 ans[cnt] = a[i];  69 vis[a[i]] = 1;  70 DFS (cnt + 1);  71 vis[a[i]] = 0;  72  }  73  }  74 }  75  76 int main(void) //ZOJ 3861 Valid Pattern Lock  77 {  78 //freopen ("B.in", "r", stdin);  79  80 int t;  81 scanf ("%d", &t);  82 while (t--)  83  {  84 //memset (res, -1, sizeof (res)); //这句话没写导致WA n次！！！  85 memset (vis, 0, sizeof (vis)); //注意初始化，上面的作用已在ok () Debug 出来了  86 scanf ("%d", &n);  87  88 for (int i=1; i<=n; ++i) scanf ("%d", &a[i]);  89 sort(a+1,a+n+1);  90 x = 0;  91 DFS (1);  92  93 printf ("%d\n", x);  94 for (int i=1; i<=x; ++i)  95  {  96 for (int j=1; j<=n; ++j)  97  {  98 printf ("%d%c", res[i][j], (j==n) ? '\n' : ' ');  99  } 100  } 101 102  } 103 104 return 0; 105 }