技術標籤：PAT甲

連結串列存在vector裡，不用再通過order來cmp來排序了
reverse(v1.begin(),v1.end())
swap(v1,v2)

#include <iostream>
#include <vector>
using namespace std;
int const maxn = 100010;

struct Node {
	int address, data, next;
	int order;
}node[maxn],node1[maxn],node2[maxn];

vector<Node>v1, v2;

int
 
 main() {

	int s1, s2, n;
	cin >> s1 >> s2 >> n;
	for (int i = 0; i < maxn;i++) {
		node[i].order= 2 * maxn;
	}

	for (int i = 0; i < n; i++) {
		int address;
		cin >> address;
		node[address].address = address;
		cin >> node[address].data >> node[address].next;
 

	}


	int p1;
	for (p1 = s1; p1 != -1; p1 = node[p1].next) {
		v1.push_back(node[p1]);
	}

	int p2;
	for (p2 = s2; p2 != -1; p2 = node[p2].next) {
		v2.push_back(node[p2]);
	}
	
	if (v1.size() < v2.size())swap(v1, v2);//v2存小的
	reverse(v2.begin(), v2.end());


	//i<2
	//v1存長的
	//第i塊
	//把短的先輸出
	for (int i =
 
 0; i < v2.size(); i++) {
		int j;
		for (j = 0; j < 2; j++) {
			if (j == 0) 
				printf("%05d %d %05d\n", v1[i * 2 + j].address, v1[i * 2 + j].data,v1[i*2 +j+1].address);
			else 
				printf("%05d %d ", v1[i * 2 + j].address, v1[i * 2 + j].data);
		}

		printf("%05d\n", v2[i].address);

		printf("%05d %d ", v2[i].address, v2[i].data);
		printf("%05d\n", v1[i * 2 + j].address);
	}
	//把剩下的長的輸出了

	for (int i = v2.size()*2;i < v1.size(); i++) {
		printf("%05d %d %d\n", v1[i].address, v1[i].data, v1[i].next);
	}

	return 0;
}
/*
00100 01000 7
02233 2 34891
00100 6 00001
34891 3 10086
01000 1 02233
00033 5 -1
10086 4 00033
00001 7 -1

01000 1 02233
02233 2 00001
00001 7 34891
34891 3 10086
10086 4 00100
00100 6 00033
00033 5 -1
*/