The 15th Zhejiang University Programming Contest B Valid Pattern Lock 构造

题意：给你 3*3的手机解锁图，问你解锁线段只穿过指定的点的种类数。

解题思路：把所有不能组成的组合打表出来特判 next_permutation就行了。

解题代码：

// File Name: b.cpp
// Author: darkdream
// Created Time: 2015年04月12日 星期日 13时51分52秒

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<ctime>
#define LL long long

using namespace std;
int t; 
int n;
int a[15];
int notok[][3] ={
    {1,3,2},{3,1,2},{1,7,4},{7,1,4},
    {1,9,5},{9,1,5},{4,6,5},{6,4,5},
    {7,9,8},{9,7,8},{7,3,5},{3,7,5},
    {2,8,5},{8,2,5},{3,9,6},{9,3,6}
};
int vis[15];
int ok()
{
    int visn[15];
    memset(visn,0,sizeof(visn));
    for(int j = 1;j <= n-1;j ++)
    {
        for(int i = 0 ;i <= 15;i ++)
        {
            if(a[j] == notok[i][0] && a[j+1] == notok[i][1] &&visn[notok[i][2]] == 0 )
                return 0 ;
        }
        visn[a[j]] = 1;
    }
    return 1; 
}
int main(){
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        memset(vis,0,sizeof(vis));
        for(int i = 1;i <= n;i ++)
        {
            scanf("%d",&a[i]);
            vis[a[i]] = 1; 
        }
        int ans = 0 ; 
        sort(a+1,a+1+n);
        do{
            if(ok())
                ans ++ ; 
        }while(next_permutation(a+1,a+1+n));
        printf("%d
",ans);
        sort(a+1,a+1+n);
        int tt = 0 ; 
        do{
            if(ok())
            {
                for(int i = 1;i <= n;i ++)
                    printf(i == 1?"%d":" %d",a[i]);
                printf("
");
                tt ++;
            }
        }while(next_permutation(a+1,a+1+n));
        //printf("%d
",tt);
        
    }
    return 0;
}