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.
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.
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=100010;
struct st
{
int address,data,next;
int order;
} node[maxn];
bool cmp(st a,st b)
{
return a.order<b.order;
}
int main()
{
for(int i=0; i<maxn; i++)
{
node[i].order=maxn;
}
int beg,n,k,address;
scanf("%d%d%d",&beg,&n,&k);
for(int i=0; i<n; i++)
{
scanf("%d",&address);
scanf("%d%d",&node[address].data,&node[address].next);
node[address].address=address;
}
int p=beg,coun=0;
while(p!=-1)
{
node[p].order=coun++;
//printf("p=%d\n",node[p].order);
p=node[p].next;
}
sort(node,node+maxn,cmp);
n=coun;
//printf("%d\n",n);
for(int i=0; i<n/k; i++)
{
for(int 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("-1\n");
}
else
{
printf("%05d\n",node[(i+1)*k].address);
for(int i=n/k*k; i<n; i++)
{
printf("%05d %d ",node[i].address,node[i].data);
if(i<n-1)
{
printf("%05d\n",node[i+1].address);
}
else
{
printf("-1\n");
}
}
}
}
}
return 0;
}
