计蒜客习题：画图游戏（Havel-Hakimi定理）

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
#define maxn 1000 + 5
struct node{
    int pos;
    int degree;
}a[maxn];
int ans[maxn][maxn];
bool cmp (node a,node b){
    return a.degree > b.degree;
}
int main()
{
    int n, x = 0;
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> a[i].degree;
        a[i].pos = i;
        if (!a[i].degree) x++;
    }
    //由Havel–Hakimi可知不能成图的条件为：
    //当所有点的度不为都0且存在点的度为0
    //这样的话降序减下去，就必然出现度为负的点
    if (x != 0 && x != n) 
    {
        cout << "None\n";
        return 0;
    }
    for (int i = 0; i < n; i++)
    {
        sort(a+i, a+n, cmp);
        if (a[i].degree==0) break;
        if(a[i].degree > n-1-i)
        {
            cout << "None";
            return 0;
        }
        for (int j = 1 + i; j <= a[i].degree + i; j++)
        {
            a[j].degree--;
            if (a[j].degree < 0)
            {
                cout << "None";
                return 0;
            }
            ans[a[i].pos][a[j].pos] = ans[a[j].pos][a[i].pos] = 1;
        }
    }
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < n; j++)
        {
            if (j == 0) cout << ans[i][j];
            else cout << " " << ans[i][j];
        }
        cout << endl;
    }
    return 0;
}