PAT（甲级）2019年冬季考试 7-2 Block Reversing

7-2 Block Reversing (25分)

Given a singly linked list L. Let us consider every K nodes as a block (if there are less than K nodes at the end of the list, the rest of the nodes are still considered as a block). Your job is to reverse all the blocks in L. For example, given L as 1→2→3→4→5→6→7→8 and K as 3, your output must be 7→8→4→5→6→1→2→3.

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 (≤10​5​​) which is the total number of nodes, and a positive K (≤N) which is the size of a block. 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 8 3
71120 7 88666
00000 4 99999
00100 1 12309
68237 6 71120
33218 3 00000
99999 5 68237
88666 8 -1
12309 2 33218`

Sample Output:

71120 7 88666
88666 8 00000
00000 4 99999
99999 5 68237
68237 6 00100
00100 1 12309
12309 2 33218
33218 3 -1

题目限制：

题目大意：

现给定一个链表L，假定每K个位一块，最后不足K个的也为一块，如今要求将全部的块进行逆置，块内保证有序。

算法思路：

PAT的静态链表的题目，要把握其中的一个特色就是，结点的地址和数值是绑定的，next彻底不须要操心，最后将全部结点放在一片内存连续的区域天然就能够获得了。咱们这里采用排序的方法来解决这个问题，给每个结点赋值一个flag表明每个结点的块号 ，id表明每个结点初始的相对顺序，那么排序规则就是先根据flag大的进行排序，flag相同的直接按照id进行排序便可。接下来就是flag和id的获取，对于id来讲，链表的起点为0，日后依次累加，而flag为其id/K，拍完序以后，直接输出便可，只须要注意next的输出，在最后一个结点得输出-1，不然输出下一个结点的地址就行。

注意点：

• 一、结点不必定都在链表上，得遍历一遍。
• 二、地址为5位整数，得保证输出和输入一个格式。

提交结果：

AC代码：

#include<cstdio>
#include<string>
#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

struct Node{
    int address;
    int data;
    int next;
    int flag;// 每个结点的块号 
    int id;// 每个结点初始的相对顺序 
}nodes[100005];
vector<Node> list;

bool cmp(const Node &a,const Node &b){
    return a.flag!=b.flag?a.flag>b.flag:a.id<b.id;
}

int main(){
    int begin,N,K;
    scanf("%d %d %d",&begin,&N,&K);
    Node node;
    for (int i = 0; i < N; ++i) {
        scanf("%d %d %d",&node.address,&node.data,&node.next);
        nodes[node.address] = node;
    }
    int num = 0;
    while(begin!=-1){
        nodes[begin].id = num;
        nodes[begin].flag = num/K;
        ++num;
        list.push_back(nodes[begin]);
        begin = nodes[begin].next;
    }
    sort(list.begin(),list.end(),cmp);
    for(int i=0;i<list.size();++i){
        printf("%05d %d ",list[i].address,list[i].data);
        if(i<list.size()-1){
            printf("%05d\n",list[i+1].address);
        } else {
            printf("-1");
        }
    }
    return 0;
}