[校内训练19_09_06]直径

题意

构造一棵树，让它有k条直径。边权都是非负整数，要求点数不能超过5000。

$k leq 5000000$

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思考

让树根底下分别挂着长度为a，b，c的边权为0的链，然后将链的某一端向根连一条边权为1的边，这样共有ab+ac+bc条直径。可以发现，对于数据范围内的k，可以很快地找到这样的a，b，c。

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代码

// luogu-judger-enable-o2
// luogu-judger-enable-o2
// luogu-judger-enable-o2
// luogu-judger-enable-o2
#include<bits/stdc++.h>
using namespace std;
typedef unsigned int ui;
ui SEED;
int k;
int size,a[5555];
int b[5000005];
bool vis[5000005];
inline int R(int l,int r)
{
    SEED^=SEED<<13;
    SEED>=SEED>>5;
    SEED^=1;
    return SEED%(r-l+1)+l;
}
int main()
{
//    freopen("diameter.in","r",stdin);
//    freopen("diameter.out","w",stdout);
    ios::sync_with_stdio(false);
    for(int i=1;i*i<=5000000;++i)
        vis[i*i]=1;
    cin>>k;
    if(k<=3000)
    {
        cout<<k+2<<endl;
        cout<<k+2<<" "<<1<<" "<<100<<endl;
        for(int i=1;i<=k;++i)
            cout<<1<<" "<<i+1<<" "<<1<<endl;
        return 0;
    }
    int base=k;
    for(int c=sqrt(base);c>=1;--c)
    {
        int g=k-c*c;
        for(int a=1;a*a<=g;++a)
            for(int b=1;b*b<=g;++b)
                if(a*b+a*c+b*c==k)
                {
                    --a,--b,--c;
                    cout<<a+b+c+4<<endl;
                    cout<<1<<" "<<2<<" "<<1<<endl;
                    for(int i=1;i<=a;++i)
                        cout<<1+i<<" "<<1+i+1<<" "<<0<<endl;
                    cout<<1<<" "<<2+a+1<<" "<<1<<endl;
                    for(int i=1;i<=b;++i)
                        cout<<2+a+i<<" "<<2+a+i+1<<" "<<0<<endl;
                    cout<<1<<" "<<3+a+b+1<<" "<<1<<endl;
                    for(int i=1;i<=c;++i)
                        cout<<3+a+b+i<<" "<<3+a+b+i+1<<" "<<0<<endl;
                    return 0;
                }
    }
    return 0;
}