Codeforces Round #285 (Div. 二) C. Misha and Forest 树

Codeforces Round #285 (Div. 2) C. Misha and Forest 树

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
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，给出所有结点相邻结点的个数与相邻结点的异或和，要求出所有的边。

由于，叶子结点只与父结点相连，所以，只要知道了叶子结点，就知道了，其父结点，这样，从叶子结点起自下而上，就可以推出所有的边。

#define N 100005
#define M 100005
#define maxn 205
#define MOD 1000000000000000007
int n,a[N],b[N],ansNum;
pii ans[N];
vector<int> p[N];
queue<pii> q;
int main()
{
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
     while(S(n)!=EOF)
    {
        while(!q.empty()) q.pop();
        FI(n){
            S2(a[i],b[i]);

        }
        FI(n){
            if(a[i] == 1){
                q.push(mp(b[i],i));
            }
        }
        ansNum = 0;
        while(!q.empty()){
            pii t = q.front();q.pop();
            if(a[t.first] >= 1){
                ans[ansNum++] = mp(t.first,t.second);
                a[t.second]--;
                b[t.first] ^= t.second;
                a[t.first]--;
                if(a[t.first] == 1)
                    q.push(mp(b[t.first],t.first));
            }
        }
        printf("%d\n",ansNum);
        FI(ansNum){
            printf("%d %d\n",ans[i].first,ans[i].second);
        }
    }
    //fclose(stdin);
    //fclose(stdout);
    return 0;
}

Codeforces Round #285 (Div. 二) C. Misha and Forest 树

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