1097. Deduplication on a Linked List (25)

时间限制 300 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue

Given a singly linked list L with integer keys, you are supposed to remove the nodes with duplicated absolute values of the keys. That is, for each value K, only the first node of which the value or absolute value of its key equals K will be kept. At the mean time, all the removed nodes must be kept in a separate list. For example, given L being 21→-15→-15→-7→15, you must output 21→-15→-7, and the removed list -15→15.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, and a positive N (<= 10⁵) which is the total number of nodes. 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 Key Next

where Address is the position of the node, Key is an integer of which absolute value is no more than 10⁴, and Next is the position of the next node.

Output Specification:

For each case, output the resulting linked list first, then the removed list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input: 00100 5 99999 -7 87654 23854 -15 00000 87654 15 -1 00000 -15 99999 00100 21 23854 Sample Output: 00100 21 23854 23854 -15 99999 99999 -7 -1 00000 -15 87654 87654 15 -1

题意：给你一个链表，每个结点数据绝对值重复的结点，仅保留第一个，其余的按地址连接的顺序去除，并且连接成另一个链表

解题思路：可以遍历一遍链表，可以用两个数组分别记录两个链表的地址 

#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <cmath> #include <map> #include <set> #include <stack> #include <queue> #include <vector> #include <bitset> using namespace std; #define LL long long const int INF = 0x3f3f3f3f; struct node { int val, nt; }a[100005]; int vis[100005], ans1[100005], ans2[100005]; int main() { int x, n; while (~scanf("%d%d", &x, &n)) { int k,cnt1=0,cnt2=0; memset(vis, 0, sizeof vis); for (int i = 1; i <= n; i++) scanf("%d", &k), scanf("%d%d", &a[k].val,&a[k].nt); for (int i = x; ~i; i = a[i].nt) { k = abs(a[i].val); if (!vis[k]) vis[k] = 1, ans1[cnt1++] = i; else ans2[cnt2++] = i; } printf("d %d ", ans1[0], a[ans1[0]].val); for (int i = 1; i<cnt1; i++) { printf("d\n", ans1[i]); printf("d %d ", ans1[i], a[ans1[i]].val); } printf("-1\n"); if (cnt2) { printf("d %d ", ans2[0], a[ans2[0]].val); for (int i = 1; i<cnt2; i++) { printf("d\n", ans2[i]); printf("d %d ", ans2[i], a[ans2[i]].val); } printf("-1\n"); } } return 0; }