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;
}