02-线性结构3 Reversing Linked List（25 分)

02-线性结构3 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 $N(\le 10^5)$ 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 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

---

#include <stdlib.h>
#include <stdio.h>


typedef struct node {
	int Address;
	int Data;
	int Next;
}list;

void Array(list a[], int N, int headaddress);
void Exchange(list a[], int i, int j);
void Reverse(list a[], int N, int K);

int main(){
	int K, N, headaddress, eff;
	scanf("%d %d %d", &headaddress, &N, &K);
	list a[N];
	for (int i = 0; i<N; i++) {
		scanf("%d %d %d", &a[i].Address, &a[i].Data, &a[i].Next);
	}
	Array(a, N, headaddress);
	for (int i = 0; i<N; i++) {
		if (a[i].Next == -1) {
			eff = i + 1;
		}
	}
	Reverse(a, eff, K);
	for (int i = 0; i<eff; i++)	{
		a[i].Next = a[i + 1].Address;
		if (i == eff - 1) {
			a[i].Next = -1;
		}
	}
	for (int i = 0; i<eff - 1; i++) {
		printf("%05d %d %05d\n", a[i].Address, a[i].Data, a[i].Next);
	}
	printf("%05d %d %d\n", a[eff - 1].Address, a[eff - 1].Data, -1);
	return 0;
}


void Array(list a[], int N, int headaddress) {
	for (int i = 0; i<N; i++) {
		if (a[i].Address == headaddress) {
			Exchange(a, i, 0);
			break;
		}
	}
	for (int i = 0; i<N; i++) {
		for (int j = i + 1; j<N; j++) {
			if (a[i].Next == a[j].Address) {
				Exchange(a, i + 1, j);
				break;
			}
		}
	}
}


void Exchange(list a[], int i, int j) {
	list temp;
	temp.Address = a[j].Address;
	temp.Data = a[j].Data;
	temp.Next = a[j].Next;
	a[j].Address = a[i].Address;
	a[j].Data = a[i].Data;
	a[j].Next = a[i].Next;
	a[i].Address = temp.Address;
	a[i].Data = temp.Data;
	a[i].Next = temp.Next;
}

void Reverse(list a[], int N, int K) {
	int cnt = N / K;
	if (cnt) {
		for (int j = 1; j <= cnt; j++) {
			for (int i = 1; i <= K; i++) {
				if ((j - 1)*K + i - 1 <= j * K - i - 1) {
					Exchange(a, (j - 1)*K + i - 1, j*K - i);
				}
			}
		}
	}
}