BNUOJ 34981 A Matrix

BNUOJ 34981 A Matrix

Address: BNUOJ 34981

Question:
Here is an algorithm and matrix for placing an arrangement in a matrix. You need to write an arrangement from the matrix.
If there are multiple answers, the largest Lexicographic Order after the flip is output.

Analysis:
After thinking about it for a long time, I can only think of the Linked List version, and it is very likely that TLE, after reading the question of Fan Shen, suddenly become clear... Orz ..

Code:

/**  Author:      illuz 
 
  *  File:        a.cpp*  Create Date: 2014-05-29 21:05:38*  Descripton:   */#include 
  
   #include 
   
    using namespace std;const int N = 1e5 + 10;int t, n, m, p[N], tmp, ans[N], cnt;bool flag;vector
    
      v[N];void dfs(int r, int val) {if (r >= m || p[r] - 1 < 0 || v[r][p[r] - 1] < val)return;ans[cnt++] = v[r][p[r] - 1];dfs(r + 1, v[r][p[r] - 1]);p[r]--;}int main(){scanf("%d", &t);for (int cas = 1; cas <= t; cas++) {printf("Case #%d:", cas);scanf("%d%d", &n, &m);flag = false;for (int i = 0; i < m; i++) {v[i].clear();scanf("%d", &p[i]);for (int j = 0; j < p[i]; j++) {scanf("%d", &tmp);v[i].push_back(tmp);if (j > 0 && v[i][j] < v[i][j - 1])flag = true;}if (i > 0 && v[i][0] < v[i - 1][0])flag = true;}if (flag) {puts(" No solution");continue;}cnt = 0;for (int i = p[0] - 1; i >= 0; i--) {ans[cnt++] = v[0][i];dfs(1, v[0][i]);}if (cnt != n) {puts(" No solution");continue;}for (int i = cnt - 1; i >= 0; i--)printf(" %d", ans[i]);puts("");}return 0;}