501C Misha and Forest (队列)

C. 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, degree_v and s_v, 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 degree_v and s_v 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 degree_i and s_i (0 ≤ degree_i ≤ n - 1, 0 ≤ s_i < 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：顺序给你n个顶点信息degree表示该点所连的边数,s表示所连各点一起异或的值,问你n条边的信息。

思路：突破口是叶子节点,再利用队列处理。

代码：最近想多练练STL,就写繁琐了点。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<queue>
using namespace std;
const int N = 100000;
queue<int>que;
vector<pair<int,int> >road;
bool vis[N];
int de[N],s[N];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        while(!que.empty())que.pop();
        road.clear();
        memset(vis,false,sizeof(vis));
        for(int i=0; i<n; i++)
        {
            scanf("%d%d",&de[i],&s[i]);
            if(de[i]==1)
                que.push(i),vis[i]=true;
        }

        while(!que.empty())
        {
            int u=que.front();
            que.pop();
            if(de[u]==1)
            {
                road.push_back(make_pair(u,s[u]));
                de[u]--;
                de[s[u]]--;
                s[s[u]]^=u;
            }
            if(!vis[s[u]])
            {
                if(de[s[u]]==1)
                    que.push(s[u]),vis[s[u]]=true;
            }
        }
        printf("%d\n",road.size());
        for(int i=0;i<road.size();i++)
            printf("%d %d\n",road[i].first,road[i].second);
    }
    return 0;
}