ZOJ 3861 Valid Pattern Lock
来源:互联网 发布:淘宝达人如何合作 编辑:程序博客网 时间:2024/05/21 12:34
Pattern lock security is generally used in Android handsets instead of a password. The pattern lock can be set by joining points on a 3 × 3 matrix in a chosen order. The points of the matrix are registered in a numbered order starting with 1 in the upper left corner and ending with 9 in the bottom right corner.
A valid pattern has the following properties:
- A pattern can be represented using the sequence of points which it's touching for the first time (in the same order of drawing the pattern). And we call those points as active points.
- For every two consecutive points A and B in the pattern representation, if the line segment connecting A and B passes through some other points, these points must be in the sequence also and comes before A and B, otherwise the pattern will be invalid.
- In the pattern representation we don't mention the same point more than once, even if the pattern will touch this point again through another valid segment, and each segment in the pattern must be going from a point to another point which the pattern didn't touch before and it might go through some points which already appeared in the pattern.
Now you are given n active points, you need to find the number of valid pattern locks formed from those active points.
Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
The first line contains an integer n (3 ≤ n ≤ 9), indicating the number of active points. The second line contains n distinct integers a1, a2, … an (1 ≤ ai ≤ 9) which denotes the identifier of the active points.
Output
For each test case, print a line containing an integer m, indicating the number of valid pattern lock.
In the next m lines, each contains n integers, indicating an valid pattern lock sequence. The m sequences should be listed in lexicographical order.
Sample Input
131 2 3
Sample Output
41 2 32 1 32 3 13 2 1
连接的可能数
#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int t,a[15],b[15],c[15][15],f[15],n,ans[500000][15],len;void print(){ int i; len++; for (i=1; i<=n; i++) ans[len][i]=f[i];}void dfs(int l){ int i; if(l>n) print(); else { for(i = 1; i<=9; i++) { if ((a[i]==1) && (b[i]==1) && (a[c[i][f[l-1]]]==0) && (b[c[i][f[l-1]]]==1)) { a[i]=0; f[l]=i; dfs(l+1); a[i]=1; f[l]=0; } } }}///(a[c[i][f[l-1]]]==0) 如果I 和 F[L-1]对应不能出现的数己经用了///(b[c[i][f[l-1]]]==1) 如果这个数存在的话int main(){ int t,i,x,j; memset(c,0,sizeof(c)); c[1][3]=2,c[3][1]=2; c[1][7]=4,c[7][1]=4; c[1][9]=5,c[9][1]=5; c[2][8]=5,c[8][2]=5; c[3][9]=6,c[9][3]=6; c[3][7]=5,c[7][3]=5; c[4][6]=5,c[6][4]=5; c[7][9]=8,c[9][7]=8; scanf("%d",&t); while (t>0) { t--; memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); b[0]=1; scanf("%d",&n); for (i=1; i<=n; i++) { scanf("%d",&x); a[x]=1; ///X有没有使用 b[x]=1; ///X能否使用 } len=0; dfs(1); printf("%d\n",len); for (i=1; i<=len; i++) { for (j=1; j<n; j++) printf("%d ",ans[i][j]); printf("%d\n",ans[i][n]); } } return 0;}
- ZOJ 3861 Valid Pattern Lock
- ZOJ 3861 Valid Pattern Lock
- zoj 3861 Valid Pattern Lock
- ZOJ 3861 Valid Pattern Lock
- ZOJ 3861 Valid Pattern Lock
- zoj 3861 Valid Pattern Lock
- ZOJ 3861 - Valid Pattern Lock
- zoj 3861 Valid Pattern Lock (DFS)
- ZOJ 3861 Valid Pattern Lock DFS
- [水题+dfs] zoj 3861 Valid Pattern Lock
- zoj 3861 Valid Pattern Lock(dfs)
- ZOJ--(5464)Valid Pattern Lock
- zoj 3861 Valid Pattern Lock 手势密码 DFS
- Valid Pattern Lock-ZOJ 暴力DFS搜索
- ZOJ - 3861 Valid Pattern Lock(dfs或其他,两种解法)
- zoj 3861 Valid Pattern Lock(以及自己对dfs的一些理解)
- ZOJ 3861 Valid Pattern Lock(深度优先遍历dfs,有限制条件的全排列)
- (stl)Valid Pattern Lock
- UISegmentControl的使用
- Java性能优化(10):谨慎改写clone
- OK6410裸机开发之Ubuntu上Openocd环境搭建
- android studio导入项目出现Gradle DSL method not found: ‘android()’
- HOU211学习总结
- ZOJ 3861 Valid Pattern Lock
- 高薪族是如何与我们一点一点和我们划清界限的
- SUID,SGID,SBIT
- 关于 U-BOOT 中 SPL 的移植——实现函数
- js控制滚动条联动
- CCF 201409-3字符串匹配 (KMP)
- C 语言中的exit函数
- leetcode: (144) Binary Tree Preorder Traversal
- 可能是ThinkPHP导航高亮显示当前页面的最简便的方法