Off white cow month season 23 G trees sum solution to a problem (greedy)

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In fact, this idea that they have a problem, but in the end did not write the code but it is not to see a solution to a problem. This question and one question is actually very similar to race weeks are seeking contributions

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This problem requires thinking in terms of contribution to count, for each edge of the contribution of the answer, which is equal to the left subtree node point number × the right subtree side length × number of nodes.

So we just from the first point DFS, the number of nodes pretreatment each subtree. Then traverse all sides.

The number left to the right subtree node × number of sub-tree nodes point product preserved. Descending order, from front to back and then again traversed respectively multiplied by 1/2/3 / ... / n-1, can be combined.

Time complexity of O (n × log (n)).

To find the nodes from around the size of the sort, because inside there are n total, but in fact only n-1 edges. There is a s [1] = 0, will certainly be discarded.

pass: two-way side do not open the screen twice to eat

Code

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=1e5+5;
ll ans,s[maxn],pos[maxn];
int head[maxn],cnt,cnt2,n,u,v;
struct node{
	int to,next;
}e[maxn<<1];//双向边 
void add(int u,int v){
	e[++cnt].to=v;
	e[cnt].next=head[u];
	head[u]=cnt;
}
void dfs(int son,int fa){
	pos[son]=1;//左边有多少个节点 
	for(int i=head[son];i;i=e[i].next){
		if(e[i].to!=fa){
			dfs(e[i].to,son);
			pos[son]=pos[son]+pos[e[i].to];
		}
	}
	s[++cnt2]=pos[son]*(n-pos[son]);//左节点数量乘以右节点数量 
}
bool cmp(ll a,ll b){
	return a>b;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n-1;i++){
		scanf("%d %d",&u,&v);
		add(u,v),add(v,u);
	}
	dfs(1,0);
	sort(s+1,s+1+cnt2,cmp);//要从大到小因为里面第一个元素为0 
	for(int i=1;i<=n-1;i++){
		ans=ans+s[i]*i;	
	}
	printf("%lld",ans);
    return 0;
}