Educational Codeforces Round 43 (Rated for Div. 2) D. Degree Set

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最新推荐文章于 2018-07-29 15:12:56 发布

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本文介绍了一种算法，该算法能够根据给定的递增正整数序列（度数集），构造一个无向图。确保图中不存在自环或多条边，并且总的边数不超过106。同时，文章提供了完整的C++实现代码。

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You are given a sequence of n positive integers d1, d2, ..., dn (d1 < d2 < ... < dn). Your task is to construct an undirected graph such that:

there are exactly dn + 1 vertices;
there are no self-loops;
there are no multiple edges;
there are no more than 106 edges;
its degree set is equal to d.

Vertices should be numbered 1 through (dn + 1).

Degree sequence is an array a with length equal to the number of vertices in a graph such that ai is the number of vertices adjacent to i-th vertex.

Degree set is a sorted in increasing order sequence of all distinct values from the degree sequence.

It is guaranteed that there exists such a graph that all the conditions hold, and it contains no more than 106 edges.

Print the resulting graph.

Input

The first line contains one integer n (1 ≤ n ≤ 300) — the size of the degree set.

The second line contains n integers d1, d2, ..., dn (1 ≤ di ≤ 1000, d1 < d2 < ... < dn) — the degree set.

Output

In the first line print one integer m (1 ≤ m ≤ 106) — the number of edges in the resulting graph. It is guaranteed that there exists such a graph that all the conditions hold and it contains no more than 106 edges.

Each of the next m lines should contain two integers vi and ui (1 ≤ vi, ui ≤ dn + 1) — the description of the i-th edge.

#include<bits/stdc++.h>
using namespace std;
const int maxn=303;
int a[maxn],tol,last,cnt;
stack<int>P,Q;
void add(int x)
{
    for(int i=x+1;i<=tol;i++)
    {
        P.push(x);
        Q.push(i);
    }
}
int main()
{
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&a[i]);
    tol=a[n]+1,last=0,cnt=0;
    int i,l,r;
    for(i=1,l=1,r=n;i<=tol&&l<=r;i++)
    {
        if(cnt+tol-i==a[r])
        {
            cnt++;r--;
            add(i);last=1;
        }
        else if(cnt==a[l]) l++,last=0;
        else if(last) cnt++,add(i);
    }
    if(last)
    {
        for( ;i<=tol;i++)add(i);
    }
    int T=Q.size();
    printf("%d\n",T);
    while(T--)
    {
        printf("%d %d\n",P.top(),Q.top());
        P.pop();Q.pop();
    }
    return 0;
}