1074 Reversing Linked List (25分)(链表区间反转)
1074 Reversing Linked List (25分)
Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K=3, then you must output 3→2→1→6→5→4; if K=4, you must output 4→3→2→1→5→6.
Input Specification:
Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (≤) which is the total number of nodes, and a positive K (≤) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.
Then N lines follow, each describes a node in the format:
Address Data Next
where Address
is the position of the node, Data
is an integer, and Next
is the position of the next node.
Output Specification:
For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.
Sample Input:
00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218
Sample Output:
00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1
题目说给一个单链,好家伙,最后一个样例里面有些节点不存在于这条单链上,真的一点都不严谨啊。
就好像题目说给你颗二叉树,结果最后整个图出现多个连通块一样。吐吐的。
1 #include <cstdio> 2 using namespace std; 3 4 const int maxn = 1e6 + 5, maxm = 1e5 + 5; 5 6 struct Linked { 7 int pre, next, data; 8 } lists[maxn]; 9 int val[maxm], cnt; 10 11 int main() { 12 int head, n, k, u, v, data; 13 scanf("%d %d %d", &head, &n, &k); 14 for(int i = 0; i < n; i ++) { 15 scanf("%d %d %d", &u, &data, &v); 16 lists[u].next = v; 17 lists[v].pre = u; 18 lists[u].data = data; 19 if(u == head) lists[u].pre = -1; 20 } 21 while(~head) { 22 Linked temp = lists[head]; 23 val[++ cnt] = head; 24 head = temp.next; 25 } 26 if(cnt >= k) { 27 for(int i = 1; i <= cnt / k; i ++) { 28 for(int j = i * k; j > (i - 1) * k + 1; j --) { 29 Linked temp = lists[val[j]]; 30 printf("%05d %d %05d\n", val[j], temp.data,temp.pre); 31 } 32 Linked temp = lists[val[k * (i - 1) + 1]]; 33 if(i < cnt / k) printf("%05d %d %05d\n", val[k * (i - 1) + 1], temp.data, val[k * (i + 1)]); 34 else { 35 if(cnt % k) { 36 printf("%05d %d %05d\n", val[k * (i - 1) + 1], temp.data, val[k * i + 1]); 37 int flag = (cnt / k) * k; 38 for(int j = 1; j < cnt % k; j ++) { 39 Linked temp = lists[val[flag + j]]; 40 printf("%05d %d %05d\n", val[flag + j], temp.data, temp.next); 41 } 42 printf("%05d %d -1\n", val[flag + cnt % k], lists[val[flag + cnt % k]].data); 43 } else { 44 printf("%05d %d -1\n", val[k * (i - 1) + 1], temp.data); 45 } 46 } 47 } 48 } else { 49 for(int i = 1; i < cnt; i ++) { 50 printf("%05d %d %05d\n", val[i], lists[val[i]].data, lists[val[i]].next); 51 } 52 printf("%05d %d -1\n", val[cnt], lists[val[cnt]].data); 53 } 54 return 0; 55 }