PAT (Advanced Level) Practise 1133 Splitting A Linked List (25)

1133. Splitting A Linked List (25)

时间限制

400 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, 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<stdio.h>
const int N = 1e5 + 10;
int nt
, u
, v
;
int x, y, z;
int s, n, m, t = -1;

int main(){
scanf("%d%d%d",&s,&n,&m);
for (int i = 1;i<=n;i++) {
scanf("%d%d%d",&x,&y,&z);
u[x] = y; nt[x] = z;
}
int now = -1;
for (int i = s;i!=-1;i = nt[i]) {
if (u[i] < 0) {
if (t < 0) t = now = i;
else {
v[now] = i; now = i;
}
}
}
for (int i = s;i!=-1;i = nt[i]) {
if (u[i] >= 0 && u[i] <= m) {
if (t < 0) t = now = i;
else {
v[now] = i; now = i;
}
}
}
for (int i = s;i!=-1;i = nt[i]) {
if (u[i] > m) {
if (t < 0) t = now = i;
else {
v[now] = i; now = i;
}
}
}
v[now] = -1;
for (int i = t;i!=-1;i=v[i]) {
printf("%05d %d ",i,u[i]);
if (v[i] == -1) printf("-1\n");
else printf("%05d\n",v[i]);
}
return 0;
}