After catching Jerryfish, Spongebob and Patrick thought they were happy enough so they went home. And then they found that Squidward had come home already :)
Squidward told them he need to drum up a marching band fast(for some reason). So he needed their help. Hence Spongebob and Patrick joined Squidward's band.
Today is their practicing day. Squidward is teaching all the members how to move at a performance. At a performance there are N rows. And in the i-th row, there are szi people. Note row may be empty(i.e. szi = 0). Everyone plays exactly one musical instrument. The musical instrument played by the j-th people in the i-th row is represented by a lower case letter cij.
Squidward ask them to move Q times. In each move, there are three types of moving:
Now it’s your turn! You need to help them to move and tell Squidward the final formation in the end.
The first line contains one integer N (1 ≤ N ≤ 105) – the number of rows.
Then N lines follow. First the i+1-th line contains one integer szi – the number of people in the i-th row. If there are some people in the row, a space character and szi characters cij are followed where cij is the musical instrument played by the j-th people in the i-th row. The sum of all the szi is less than or equal to 106.
The N+2-th line contains one integer Q (1 ≤ Q ≤ 105) – the times they need to move.
Then Q lines follow. The i+N+3-th line contains three integer typei, ai, bi (1 ≤ typei ≤ 3, 1 ≤ ai, bi ≤ N, ai ≠ bi) – the type of moving and the numbers of two rows which need to move.
It's guaranteed that:
Hint:
If you have no idea how to implement, please take it as your reference.
#include <stdlib.h> #include <stdio.h> typedef struct _Node { char val; struct _Node* next; } Node; //list[i]'s head node Node *head[100005] = {}; //list[i]'s tail node Node *tail[100005] = {}; void swap(int a, int b) { //swap list[a] and list[b] Node *tmp = (Node *)malloc(sizeof(Node)); /* swap(head_node) tmp = head[a]; head[a] = head[b]; head[b] = tmp; */ /* swap(tail_node) */ } void append(int a, int b) { //append list[a] to list[b]'s behind if(head[a] == NULL) return; if(head[b] == NULL) { swap(a, b); return; } /* tail[b]->next = head[a]; tail[b] = ... ... */ }
After all the moving, for each row output all the musical instrument played in that row and then print a newline('\n') character.
If there is no people in some row, just output an empty line.