codeforces round 285 div285 Misha and Forest

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,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

32 31 01 0

output

21 02 0

input

21 11 0

output

10 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".

原题链接：http://codeforces.com/contest/501/problem/C

题目大意：

输入数据是个森林，要你输出所有的边。

变相的拓扑排序。。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <set>#include <queue>#include <vector>using namespace std;#define maxn 70000#define inf 0x7ffffffint n;struct Node{    int index;    int de;    int sum;};Node arr[maxn];queue<Node> q;int main(){    scanf("%d",&n);    int degree = 0;    while(!q.empty()){        q.pop();    }    for(int i = 0; i < n; i++){        int x,y;        scanf("%d%d",&x,&y);        arr[i].index = i;        arr[i].de = x;        arr[i].sum = y;        degree += x;        if(x == 1){            q.push(arr[i]);        }    }    printf("%d\n",degree/2);    while(!q.empty()){        Node tmp = q.front();        q.pop();        arr[tmp.index].de = 0;        if(arr[tmp.sum].de != 0){            printf("%d %d\n",tmp.index,tmp.sum);        }        arr[tmp.sum].de--;        arr[tmp.sum].sum ^= tmp.index;        if(arr[tmp.sum].de == 1){            q.push(arr[tmp.sum]);        }    }    return 0;}