ZOJ - 3861 Valid Pattern Lock 【全排列】

题目链接

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3861

思路

先生成全排列，然后判断哪些情况不符合的，剔除就好了

代码中 init（） 部分 就是 先指明 哪两个数字之间 是必须有另外一个数字的

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a) memset(a, 0, sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair<string, int> psi;
typedef pair<string, string> pss;

const double PI = acos(-1);
const double E = exp(1);
const double eps = 1e-30;

const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
const int MOD = 1e9 + 7;

int n;

int c[10][10];

int vis[10];

bool judge(vector <int> ans)
{
    for (int i = 0; i < n - 1; i++)
    {
        if (c[ans[i]][ans[i + 1]] && vis[c[ans[i]][ans[i + 1]]] == 0)
            return false;
        vis[ans[i]] = 1;
    }
    return true;
}

void init()
{
    CLR(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;
}

int main()
{
    init();
    int t;
    cin >> t;
    while (t--)
    {
        scanf("%d", &n);
        vector <int> pre;
        int num;
        for (int i = 0; i < n; i++)
        {
            scanf("%d", &num);
            pre.pb(num);
        }
        sort(pre.begin(), pre.end());
        vector < vector <int> > ans;
        do
        {
            CLR(vis);
            if (judge(pre))
                ans.pb(pre);
        } while (next_permutation(pre.begin(), pre.end()));
        int len = ans.size();
        printf("%d\n", len);
        for (int i = 0; i < len; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (j)
                    printf(" ");
                printf("%d", ans[i][j]);
            }
            printf("\n");
        }
    }
}