题目链接：

题意：

给四个不同点a,b,c,d，求是否能构造出两条哈密顿通路，一条a到b，一条c到d。

题解：

构造法，看例子：

input：

5 6

5 2 4 1

output:

5 4 3 1 2

4 5 3 2 1

所以只要满足k>=n+1,就可以构造出来答案。

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          using namespace std;int n, m;const int maxn = 1111;int arr[maxn],mp[maxn];int arr2[maxn];int main() {    int a, b, c, d;    while (scanf("%d%d", &n, &m) == 2 && n) {        scanf("%d%d%d%d", &a, &b, &c, &d);        if (n ==4 || m < n + 1) {            printf("-1\n"); continue;        }        memset(mp, 0, sizeof(mp));        memset(arr, 0, sizeof(arr));        mp[a] = mp[b] = mp[c] = mp[d] = 1;        arr[1] = a; arr[n] = b;        arr[2] = c; arr[4] = d;        int tot = 5;        for (int i = 1; i <= n; i++) {            if (mp[i]) continue;            if (arr[3] == 0) arr[3] = i;            else {                arr[tot++]= i;            }        }        for (int i = 1; i < n; i++) printf("%d ", arr[i]); printf("%d\n", arr[n]);        memset(arr2, 0, sizeof(arr2));        arr2[1] = c; arr2[n] = d; arr2[2] = arr[1]; arr2[3] = arr[3];        tot = 4;        for (int i = n; i >=5; i--) arr2[tot++] = arr[i];        for (int i = 1; i < n; i++) printf("%d ", arr2[i]); printf("%d\n", arr2[n]);    }    return 0;}