标签：bar element nowrap first item bsp specific form lists
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.
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.
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.
00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218
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 }
1074 Reversing Linked List (25分)(链表区间反转)
标签：bar element nowrap first item bsp specific form lists
原文地址：https://www.cnblogs.com/bianjunting/p/13111157.html
