Codeforces Round #423 (Div. 1, rated, based on VK Cup Finals) B HIGH ROAD
2017-07-12 14:13
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构造题。题目大意:构造一棵有N个节点,其中K个是孩子节点的树,让它的直径最小,并输出这棵树的所有边和直径。
不难发现,孩子节点越多,这棵树的直径就越小。那么我们考虑删去K个孩子节点后的树。这棵树的孩子节点数肯定不能超过K,所以就让它的孩子节点数为K。一直进行下去,我们就得到了一棵每片花瓣长度为K或K-1的菊花树。那么这样构造出
4000
来的树就能满足直径最小。
那么直径是多少呢?你可以跑出来,当然也可以直接算。这里我就不赘述了,大家看我代码吧。
不难发现,孩子节点越多,这棵树的直径就越小。那么我们考虑删去K个孩子节点后的树。这棵树的孩子节点数肯定不能超过K,所以就让它的孩子节点数为K。一直进行下去,我们就得到了一棵每片花瓣长度为K或K-1的菊花树。那么这样构造出
4000
来的树就能满足直径最小。
那么直径是多少呢?你可以跑出来,当然也可以直接算。这里我就不赘述了,大家看我代码吧。
#include<bits/stdc++.h> int n,k,sc[200100]; int main(){ scanf("%d%d",&n,&k); for(int i=0;i<k;i++) sc[i]=1; printf("%d\n",(n-2+k)/k*2-((n-1)%k==1)); for(int i=2;i<=n;i+=k){ for(int j=0;j<k&&j+i<=n;j++){ printf("%d %d\n",sc[j],i+j); sc[j]=i+j; } } return 0; }
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