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

 

 题解：（关键点）为节点设立order计数，方便排序，(复杂点)每一块的处理

最后一个测试点过不去，难道是算法笔记上代码有问题？柳神代码正确

#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
",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
",node[(i+2)*k-1].address);
        }
        else{
            if(n%k==0){
                printf("%d
",-1);
            }
            else
            {
                printf("%05d
",node[(i+1)*k].address);
                for(j=n/k*k;j<n;j++){
                    if(j!=n-1){
                        printf("%05d %d %05d
",node[j].address,node[j].data,node[j+1].address);
                    }
                    else{
                        printf("%05d %d %d
",node[j].address,node[j].data,-1);
                    }
                    
                }
                    
            }

    }
    }
    return 0;
}