1074 Reversing Linked List (25 分)
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 (≤) which is the total number of nodes, and a positive K (≤) 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<bits/stdc++.h> using namespace std; const int maxn=1000010; #define inf 0x3fffffff struct Node{ int address,data,next; int order; }node[maxn]; bool cmp(Node a,Node b){ return a.order<b.order; } int main(){ for(int i=0;i<maxn;i++){ node[i].order=maxn; } int begin,n,k; scanf("%d %d %d",&begin,&n,&k); int count=0; int address,data,next; for(int i=0;i<n;i++){ scanf("%d %d %d",&address,&data,&next); node[address].address=address; node[address].data=data; node[address].next=next; } int p=begin; while(p!=-1){ node[p].order=count++; p=node[p].next; } sort(node,node+maxn,cmp); int i,j; for(i=0;i<count/k;i++){ for(j=(i+1)*k-1;j>i*k;j--){ printf("%05d %d %05d\n",node[j].address,node[j].data,node[j-1].address); } printf("%05d %d ",node[i*k].address,node[i*k].data); if(i<n/k-1){ printf("%05d\n",node[(i+2)*k-1].address); } else{ if(n%k==0){ printf("%d\n",-1); } else { printf("%05d\n",node[(i+1)*k].address); for(j=n/k*k;j<n;j++){ if(j!=n-1){ printf("%05d %d %05d\n",node[j].address,node[j].data,node[j+1].address); } else{ printf("%05d %d %d\n",node[j].address,node[j].data,-1); } } } } } return 0; }