PAT甲级1074. Reversing Linked List (25)

Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 4, you must output 4→3→2→1→5→6.

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 (<=N) which is the length of the sublist to be reversed. 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, and Next is the position of the next node.

Output Specification:

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

Sample Input:

00100 6 4

00000 4 99999

00100 1 12309

68237 6 -1

33218 3 00000

99999 5 68237

12309 2 33218

Sample Output:

00000 4 33218

33218 3 12309

12309 2 00100

00100 1 99999

99999 5 68237

68237 6 -1

#include <cstdio>
using namespace std;
#include <vector>

const int MAX=100000+10;
int node[MAX],Next[MAX],previous[MAX];

int main(){
int head,N,K;
scanf("%d %d %d",&head,&N,&K);

int Address,Data,Next_Address;
for(int i=0;i<N;i++){
scanf("%d %d %d",&Address,&Data,&Next_Address);
if(Address!=-1){
node[Address]=Data;
Next[Address]=Next_Address;
}
if(Address==head) previous[Address]=-1;
if(Next_Address!=-1) previous[Next_Address]=Address;
}

vector<int> index;
int cnt=0;//链表中结点的个数
for(int i=head;i!=-1;i=Next[i]){
index.push_back(i);
//  printf("previous=%d\n",previous[i]);
cnt++;
}

int round=cnt/K;
int m=K-1;
int last=-1;
while(m<cnt){
int i=m
4000
;
for(i;i>m-K+1;i--){
printf("%05d %d %05d\n",index[i],node[index[i]],previous[index[i]]);
}
if(m==cnt-1) printf("%05d %d %d\n",index[i],node[index[i]],-1);
else if(m+K>=cnt) printf("%05d %d %05d\n",index[i],node[index[i]],index[m+1]);
else printf("%05d %d %05d\n",index[i],node[index[i]],index[m+K]);
m+=K;
}
for(int i=m-K+1;i<cnt;i++){
if(i!=cnt-1) printf("%05d %d %05d\n",index[i],node[index[i]],Next[index[i]]);
else printf("%05d %d %d\n",index[i],node[index[i]],-1);
}

return 0;
}