Codeforces #285 C Misha and Forest 队列维护解题报告_c. misha and forest codeforce-优快云博客

本文介绍了一种基于节点度数及相邻节点异或和的森林图还原算法。通过分析节点属性，利用异或运算特性，逐步确定图中边的关系，最终还原出原始森林图的结构。

Let's define a forest as a non-directed acyclic graph (also without loops and parallel edges). One day Misha played with the forest consisting of n vertices. For each vertex v from 0 to n - 1 he wrote down two integers, degreev and sv, were the first integer is the number of vertices adjacent to vertex v, and the second integer is the XOR sum of the numbers of vertices adjacent to v (if there were no adjacent vertices, he wrote down 0).

Next day Misha couldn't remember what graph he initially had. Misha has values degreev and sv left, though. Help him find the number of edges and the edges of the initial graph. It is guaranteed that there exists a forest that corresponds to the numbers written by Misha.

Input

The first line contains integer n (1 ≤ n ≤ 216), the number of vertices in the graph.

The i-th of the next lines contains numbers degreei and si (0 ≤ degreei ≤ n - 1, 0 ≤ si < 216), separated by a space.

Output

In the first line print number m, the number of edges of the graph.

Next print m lines, each containing two distinct numbers, a and b (0 ≤ a ≤ n - 1, 0 ≤ b ≤ n - 1), corresponding to edge (a, b).

Edges can be printed in any order; vertices of the edge can also be printed in any order.

Sample test(s)

input
3
2 3
1 0
1 0

output
2
1 0
2 0

input
2
1 1
1 0

output
1
0 1

题目大意为一个森林，对于每个节点记录它的度数degree和与它连接的点的异或和Sv。现在给出每个点的degree和Sv，求这个图有多少条边，并写出每条边的顶点。

分析：

首先从异或和出发，加入x^y^z = a，那么左右都异或x的话，(x^x)^y^z = a^x，即y^z = a^x，同理z = a^x^y。

其次由于是森林，肯定有叶子节点，找出这些叶子节点，他们的degree均为1，且Sv为父节点的编号，每遍历一个叶子节点，将其父节点度数-1，Sv异或叶子节点编号，直到找到最后一个节点即可。

最后，需要用队列维护，不然会T。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <map>
using namespace std;
struct node{
	int degree,s,i;
	int vis;
	void output(){
		cout<<i<<" "<<degree<<" "<<s<<endl; 
	}
}v[70000];
typedef pair<int, int> PII;
vector<PII> edge;
vector<PII>::iterator it;
queue <int> q;
int main()
{
	freopen("in.txt","r",stdin);
	int n;
	scanf("%d",&n);
	int cnt = 0;
	for(int i = 0 ; i < n;i++)
	{
		scanf("%d%d",&v[i].degree,&v[i].s);
		if(v[i].degree == 1)
			q.push(i);
	}
	while(!q.empty())
	{
		int i = q.front();
		q.pop();
		if(v[i].degree == 1)
		{
			int fa = v[i].s;
			v[fa].s ^= i;
			v[fa].degree --;
			edge.push_back(make_pair(i,fa));
			if(v[fa].degree == 1)
				q.push(fa);				
		}	
	}

	cout<<edge.size()<<endl;
	for(it = edge.begin();it!=edge.end();it++)
	{
		PII pb = *it;
		cout<<pb.first<<" "<<pb.second<<endl;
	} 
}