PAT A1165 Block Reversing

1165 Block Reversing

分数 25

作者陈越

单位浙江大学

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 (≤105) 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

将链表用静态数组表示，将每个节点用一个结构体存储，结构体信息包括
* 节点地址add，值dat，下一个结点的地址nex，还有一个flag值，此值为了
* 让链表排序，有效节点的值从小到大排序，无效结点的值赋值为无穷大；
* 此后给数组a一个排序，就能得到链表从小到大的排列顺序；

/*** 将链表用静态数组表示，将每个节点用一个结构体存储，结构体信息包括* 节点地址add，值dat，下一个结点的地址nex，还有一个flag值，此值为了* 让链表排序，有效节点的值从小到大排序，无效结点的值赋值为无穷大；* 此后给数组a一个排序，就能得到链表从小到大的排列顺序；
*/#include <iostream>
#include <algorithm>using namespace std;struct Node
{int add, dat, nex, flag;bool operator < (const Node & a) const{return flag < a.flag;}
};const int N = 1e5+10;
struct Node a[N];
int head, n, k;void Read()
{cin >> head >> n >> k;for(int i=0; i<n; ++i){int add, dat, nex;cin >> add >> dat >> nex;a[add] = {add, dat, nex};}
}void Deal()
{for(int i=0; i<N; ++i)a[i].flag = N;int cnt = 0;while(head != -1){a[head].flag = ++cnt; //有效结点的个数head = a[head].nex;}sort(a, a+N);int block = (cnt + k - 1) / k; //链表块的块数for(int i=block-1; i>=0; --i){//如果最后一块链表不足k个，则单独处理if(i == block - 1 && cnt % k){for(int j=i*k; j<cnt; ++j){if(i == 0 && j == cnt-1) //如果是第一块的最后一个结点printf("%05d %d -1\n", a[j].add, a[j].dat);else if(j == cnt-1) //如果是最后一块的最后一个节点printf("%05d %d %05d\n", a[j].add, a[j].dat, a[(i-1)*k].add);else printf("%05d %d %05d\n", a[j].add, a[j].dat, a[j+1].add);}}else{for(int j=i*k; j<(i+1)*k; ++j){if(i == 0 && j == (i+1)*k-1) //如果是第一块的最后一个结点printf("%05d %d -1\n", a[j].add, a[j].dat);else if(j == (i+1)*k-1) //如果是最后一个节点printf("%05d %d %05d\n", a[j].add, a[j].dat, a[(i-1)*k].add);else printf("%05d %d %05d\n", a[j].add, a[j].dat, a[j+1].add);}}}
}int main()
{Read();Deal();return 0;
}