CF 1098 C. Construct a tree

题目大意：

对于一棵树，我们定义“分支系数”为子节点最多的节点的子结点个数。现给出结点个数 $n$ 、所有节点的子树大小之和 $s$ ，同时规定符合条件的树的根节点必须是 $1$ 。请你判断是否存在一棵符合条件的树。如果存在，请你输出“分支系数”最小的符合条件的树。输出方式是，你需要分别输出编号为 $2$ 的节点～编号为 $n$ 的节点的父节点。

思路：
考虑如果没有分支系数最小的限制，那么一条链的时候答案为 $i*(i+1)/2$ ，每次把最底层的叶子向上移动一层，答案就减 $1$ ，可以构造出一组解，但是答案不一定优。

显然，分支系数越大节点的子树大小之和上界越小。考虑二分出一个答案可行答案，之后在二分出的答案的基础上把节点贪心的往上层移动，可以构造出一组合法的解。

以下代码：

#include<bits/stdc++.h>
#define il inline
#define LL long long
#define _(d) while(d(isdigit(ch=getchar())))
using namespace std;
const int N=1e5+5;
vector<int> v[N];
int n,fa[N],sz[N];LL s,sum,res;
il int read(){int x,f=1;char ch;_(!)ch=='-'?f=-1:f;x=ch^48;_()x=(x<<1)+(x<<3)+(ch^48);return f*x;
}
il bool pd(int x){res=0;int t=n;LL now=1;int l=1;while(t){if(t>=now){res+=now*l;t-=now;}else{res+=t*l;t=0;}now*=x;l++;}return res<=s;
}
int main()
{n=read();scanf("%I64d",&s);if(s<n*2-1||1ll*n*(n+1)/2<s){puts("No");return 0;}if(1ll*n*(n+1)/2==s){puts("Yes");for(int i=1;i<n;i++)printf("%d ",i);return 0;}int l=1,r=n,ret=n;while(l<=r){int mid=(l+r)>>1;if(pd(mid))r=mid-1,ret=mid;else l=mid+1;}for(int i=1;i<=n;i++)v[i].push_back(i);s=1ll*n*(n+1)/2-s;for(int i=2;i<=n;i++)fa[i]=i-1,sz[i-1]=1;int now=1;for(int i=n;i;i--){while(v[now].size()==0)now++;if(i-now-1<=s){s-=(i-now-1);int k=v[now].size()-1;int d=v[now][k];sz[d]++;if(sz[d]==ret)v[now].pop_back();v[now+1].push_back(i);sz[fa[i]]--;fa[i]=d;}else{int k=i-s-1;int x=v[k][0];fa[i]=x;s=0;}if(s==0)break;}//printf("!!!%d ",ret);puts("Yes");for(int i=2;i<=n;i++)printf("%d ",fa[i]);return 0;
}

View Code

转载于:https://www.cnblogs.com/Jessie-/p/10544622.html

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CF 1098 C. Construct a tree

题目大意：

以下代码：

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