【NOIP2014 Day1 T2】联合权值

【NOIP2014 Day1 T2】联合权值

Time Limit:10000MS  Memory Limit:131072K
Total Submit:57 Accepted:30 
Case Time Limit:1000MS

Description

 

Input

 

Output

 

Sample Input

5
1 2
2 3
3 4
4 5
1 5 2 3 10

Sample Output

20 74

Hint

 

Source

感谢nodgd放上题目

看了题解才能得全分。以中间点讨论

答案一：记录n个中间点的邻近的第一大数和第二大数，更新答案

答案二：sum记录距离中间点为一的权值和, powersum记录距离中间点为一的取值平方和

中间点周围的权值和=w[1]*(sum-w[1])+w[2]*(sum-w[2])+......+w[n]*(sum-w[n])

=(w[1]+w[2]+......w[n])*sum-(w[1]*w[1]+w[2]*w[2]+......w[n]*w[n])

=sum*sum-powersum

#include<cstdio>
using namespace std;
#define maxn 200009
int u[maxn], v[maxn];
long long w[maxn], first[maxn], second[maxn], sum[maxn], powersum[maxn];
void add(int mid, long long w)
{
	sum[mid]+=w;
	powersum[mid]+=(w*w);
	if(w>first[mid])	second[mid]=first[mid], first[mid]=w;
	else if(w>second[mid])	second[mid]=w;
}
inline void _read(int& d)
{
	char t=getchar();bool f=false;
	while(t<'0'||t>'9') {if(t=='-') f=true; t=getchar();}
	for(d=0;t<='9'&&t>='0';t=getchar()) d=d*10+t-'0';
	if(f) d=-d;
}
inline void _read2(long long& d)
{
	char t=getchar();bool f=false;
	while(t<'0'||t>'9') {if(t=='-') f=true; t=getchar();}
	for(d=0;t<='9'&&t>='0';t=getchar()) d=d*10+t-'0';
	if(f) d=-d;
}
int main()
{
	int i;
	int n;
	long long ans1=0, ans2=0;
	scanf("%d", &n);
	for(i=1; i<n; i++)	_read(u[i]), _read(v[i]);
	for(i=1; i<=n; i++)	_read2(w[i]);
	for(i=1; i<=n; i++)	add(u[i], w[v[i]]), add(v[i], w[u[i]]);
	for(i=1; i<=n; i++)
	{
		if(first[i]*second[i]>ans1)	ans1=first[i]*second[i];
		ans2=((ans2+sum[i]*sum[i]-powersum[i])%10007+10007)%10007;
	}
	printf("%I64d %I64d", ans1, ans2);
}