Codeforces 501C Misha and Forest【bfs+位运算】

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.

Examples
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：
定义一个森林是一个无向无环的图，给你n个顶点（从0 - n-1）。对于每一个顶点有输入两个值，一个表示顶点的度数，一个表示与该顶点按位异或的和。（没有，就记住0）。现在要你根据这个条件把图复原出来。
思路：
对于叶子来说，只有一条边与他相连，把这条边打印，然后去掉，继续寻找叶子，直到结束。
如果一个数异或另外一个数两次，那么就变成了这个数本身。利用这个特点，将每个叶子的影响去掉。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
//typedef long long LL;
//typedef __int64 Int;
typedef pair<int, int> PAI;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
const int MAXN = (1 << 16) + 10;
PAI data[MAXN];
int main() {
    int n;
    while (scanf("%d", &n) != EOF) {
        queue <int> que;
        for (int i = 0; i < n; i++) {
            scanf("%d%d", &data[i].first, &data[i].second);
            if (data[i].first == 1) que.push(i);
        }
        PAI ans[MAXN];
        int cnt = 0;
        while (que.size()) {
            int u = que.front(); que.pop();
            if (data[u].first != 1) continue;
            int v = data[u].second; data[u].first = 0;
            ans[cnt].first = u, ans[cnt++].second = v;
            data[v].second ^= u; data[v].first--;
            if (data[v].first == 1) {
                que.push(v);
            }
        }
        printf("%d
", cnt);
        for (int i = 0; i < cnt; i++) {
            printf("%d %d
", ans[i].first, ans[i].second);
        }
    }
    return 0;
}