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");
}
}
}