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"); } } }转载于:https://www.cnblogs.com/Dup4/p/9433144.html
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