PAT 甲级 1133. Splitting A Linked List (25)

Given a singly linked list, you are supposed to rearrange its elements so that all the negative values appear before all of the non-negatives, and all the values in [0, K] appear before all those greater than K. The order of the elements inside each class must not be changed. For example, given the list being 18→7→-4→0→5→-6→10→11→-2 and K being 10, you must output -4→-6→-2→7→0→5→10→18→11.

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 (<= 105) which is the total number of nodes, and a positive K (<=1000). 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 in [-105, 105], and Next is the position of the next node. It is guaranteed that the list is not empty.

Output Specification:

For each case, output in order (from beginning to the end of the list) the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:
00100 9 10
23333 10 27777
00000 0 99999
00100 18 12309
68237 -6 23333
33218 -4 00000
48652 -2 -1
99999 5 68237
27777 11 48652
12309 7 33218
Sample Output:
33218 -4 68237
68237 -6 48652
48652 -2 12309
12309 7 00000
00000 0 99999
99999 5 23333
23333 10 00100
00100 18 27777
27777 11 -1

#include <iostream>#include <vector>using namespace std;struct node {    int data, next;}list[100000];vector<int> v[3];int main() {    int start, n, k, a;    scanf("%d%d%d", &start, &n, &k);    for (int i = 0; i < n; i++) {        scanf("%d", &a);        scanf("%d%d", &list[a].data, &list[a].next);    }    int p = start;    while (p != -1) {        int data = list[p].data;        if (data < 0)            v[0].push_back(p);        else if (data >= 0 && data <= k)            v[1].push_back(p);        else            v[2].push_back(p);        p = list[p].next;    }    int flag = 0;    for (int i = 0; i < 3; i++) {        for (int j = 0; j < v[i].size(); j++) {            if (flag == 0) {                printf("%05d %d ", v[i][j], list[v[i][j]].data);                flag = 1;            }            else {                printf("%05d\n%05d %d ", v[i][j], v[i][j], list[v[i][j]].data);            }        }    }    printf("-1");    return 0;}