Codeforces Round #285 (Div. 1) A. Misha and Forest 拓扑排序

A. Misha and Forest

time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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

Note

The XOR sum of numbers is the result of bitwise adding numbers modulo 2. This operation exists in many modern programming languages. For example, in languages C++, Java and Python it is represented as "^", and in Pascal — as "xor".

题意

给你一个森林，然后给你每个点周围有多少点，以及这个点和周围点编号的异或和，然后让你输出这些边

题解

唔，这是一道拓扑排序，不容易一样看出来……
首先，边的个数，就是相邻点数和/2
然后把入度为1的点放进队列中，然后与他相邻的那个点的坐标，就是那个异或值，然后跑一发拓扑排序就是，详情看代码吧~

代码

struct node
{
	int id;
	int x;
	int y;
};
node kiss[maxn];
int main()
{
	int n;
	cin>>n;
	int ans=0;
	REP(i,n)
	{
		kiss[i].id=i;
		cin>>kiss[i].x;
		cin>>kiss[i].y;
		ans+=kiss[i].x;
	}
	queue<node> q;
	for(int i=0;i<n;i++)
	{
		if(kiss[i].x==1)
		{
			q.push(kiss[i]);
		}
	}
	cout<<ans/2<<endl;
	while(!q.empty())
	{
		node now=q.front();
		q.pop();
		if(kiss[now.id].x!=1)
			continue;
		cout<<now.id<<" "<<now.y<<endl;
		kiss[now.id].x--;
		kiss[now.y].x--;
		kiss[now.y].y^=now.id;
		if(kiss[now.y].x==1)
			q.push(kiss[now.y]);
	}
	return 0;
}