02-线性结构3 Reversing Linked List (25分)

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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 $N$ ( $\le 10^5$ ) which is the total number of nodes, and a positive $K$ ( $\le N$ ) 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 Nextis 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 400000 4 9999900100 1 1230968237 6 -133218 3 0000099999 5 6823712309 2 33218

Sample Output:

00000 4 3321833218 3 1230912309 2 0010000100 1 9999999999 5 6823768237 6 -1
这题花了一点时间，但是并不难，第一要注意有可能不是所有的节点都要用到，所以有可能最后输出小于给出的节点。后来想到可以用reverse直接做的，然后下个节点的地址也可以不用存储，最后再输出。这个代码很粗糙，下次再来优化。
#include<iostream>#include<cstdio>using namespace std;struct node{int address;int key;int next;}a[100005],b[100005],c[100005];int main(){int ad;int m,n;cin>>ad>>n>>m;for(int i=0;i<n;i++){node A;cin>>A.address>>A.key>>A.next;a[A.address]=A;}int k=1;long temp=ad;while(a[temp].next!=-1){b[k++].address=temp;b[k-1].key=a[temp].key;b[k-1].next=a[temp].next;temp=a[temp].next;}b[k].address=a[temp].address;b[k].key=a[temp].key;b[k].next=-1;int t=0;int i;for(i=1;i*m<=k;i++){for(int j=i*m;j>(i-1)*m;j--){c[t++].address=b[j].address;c[t-1].key=b[j].key;}}for(int j=(i-1)*m+1;j<=k;j++){c[t++].address=b[j].address;c[t-1].key=b[j].key;}  for(int q=0;q<k-1;q++){c[q].next=c[q+1].address;}c[k-1].next=-1;for(int q=0;q<k-1;q++){printf("%05ld %d %05ld\n",c[q].address,c[q].key,c[q].next);}printf("%05ld %d %d\n",c[k-1].address,c[k-1].key,-1);return 0;}

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