题目链接：
题目
A. Misha and Forest
time limit per test 1 second
memory limit per test 256 megabytes
问题描述
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.
输入
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.
输出
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.
样例
input
3
2 3
1 0
1 0
output
2
1 0
2 0
题意
给你每个点的度数和所有与它直接相连的点的编号的xor的值，求这棵森林的所有的边。
题解
每次找度数为1的点，用队列维护一下，类似于拓扑排序跑一跑就可以了。
代码
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<vector>
#define X first
#define Y second
#define mp make_pair
using namespace std;
const int maxn =(1<<16)+10;
int n, m;
int deg[maxn], sum[maxn];
int main() {
scanf("%d", &n);
queue<int> Q;
for (int i = 0; i < n; i++) {
scanf("%d%d", °[i], &sum[i]);
if (deg[i] == 1) Q.push(i);
}
vector<pair<int, int> > ans;
while (!Q.empty()) {
int u = Q.front(); Q.pop();
if (deg[u] == 0) continue;
int v = sum[u];
ans.push_back(mp(v, u));
deg[v]--; sum[v] ^= u;
if (deg[v] == 1) Q.push(v);
}
printf("%d\n", ans.size());
for (int i = 0; i < ans.size(); i++) printf("%d %d\n", ans[i].X, ans[i].Y);
return 0;
}