标签：c++
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.
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.
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.
3 2 3 1 0 1 0
2 1 0 2 0
2 1 1 1 0
1 0 1
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：给出一幅无向无环图（注意无环） 共有N个点 输入每个点的 度数 和 与这点相邻的所有点的异或和
输出 图中所有的边
根据无环图 可知 图中必有度数为1的点（即只与一个顶点相邻），则它的异或和就是与它相邻顶点的下标，
这样就可以得到一条边，接着 模拟 去掉 之前度数为1的顶点 （相邻顶点度数-1，而异或和则除去去掉顶点的那一部分） 重复上述操作 即可得到所有边
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
int degree[66666],xsum[66666];
int edge[666666][2];
int main()
{
int i,j,k1,k2,n;
int x=0;
scanf("%d",&n);
queue<int>q;
for(i=0; i<n; i++)
{
scanf("%d%d",°ree[i],&xsum[i]);
if(degree[i]==1)
q.push(i);
}
while(!q.empty())
{
k1=q.front();
q.pop();
if(degree[k1]==0) //度数为0 即没有边 因为可能加入队列后 度数变为0
continue;
x++;
k2=xsum[k1]; //只有一个顶点与它相连 则 xorsum就是那个顶点下标
edge[x][0]=k1;
edge[x][1]=k2;
degree[k2]--;
if(degree[k2]==1)
q.push(k2);
xsum[k2]=xsum[k2]^k1; //xor 的精髓 a^b=c 则a^c=b b^c=a;
}
cout<<x<<endl;
for(i=1;i<=x;i++)
printf("%d %d\n",edge[i][0],edge[i][1]);
return 0;
}
codeforce #501 c Misha and Forest
标签：c++
原文地址：http://blog.csdn.net/axuan_k/article/details/42677291
