PAT (Advanced Level) Practise1133Splitting 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 1023333 10 2777700000 0 9999900100 18 1230968237 -6 2333333218 -4 0000048652 -2 -199999 5 6823727777 11 4865212309 7 33218

Sample Output:

33218 -4 6823768237 -6 4865248652 -2 1230912309 7 0000000000 0 9999999999 5 2333323333 10 0010000100 18 27777
27777 11 -1
#include<stdio.h>const int N = 1e5 + 10;int nt[N], u[N], v[N];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;}