solution

P1075的简单变形

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
struct node{
	int data, next;
}list[100000];
int main(){
	int first1, first2, n, addr;
	vector<int> l1, l2, ans;
	scanf("%d%d%d", &first1, &first2, &n);
	for(int i = 0; i < n; i++){
		scanf("%d", &addr);
		scanf("%d%d", &list[addr].data, &list[addr].next);
	}
	while(first1 != -1){
		l1.push_back(first1);
		first1 = list[first1].next;
	}
	while(first2 != -1){
		l2.push_back(first2);
		first2 = list[first2].next;
	}
	if(l2.size() > l1.size()) swap(l1, l2);
	for(int i = 0, j = l2.size() - 1; i < l1.size(); i++){
		ans.push_back(l1[i]);
		if(j >= 0 && i % 2) ans.push_back(l2[j--]); 
	}
	for(int i = 0; i < ans.size(); i++){
		if(i == 0) printf("%05d %d ", ans[i], list[ans[i]].data);
		else printf("%05d\n%05d %d ", ans[i], ans[i], list[ans[i]].data);
	}
	printf("-1");
	return 0;
}