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解题思路
直接暴力枚举每个点,然后再枚举这个点的相连点,然后用一个前缀和,复杂度不会证明,,
代码
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int MAXN = 200005;
const int mod = 10007;
typedef long long LL;
inline int rd(){
int x=0,f=1;char ch=getchar();
while(!isdigit(ch)) {f=ch=='-'?0:1;ch=getchar();}
while(isdigit(ch)) {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
return f?x:-x;
}
int n,head[MAXN],cnt;
int to[MAXN<<1],nxt[MAXN<<1];
int val[MAXN];
LL mx,ans;
inline void add(int bg,int ed){
to[++cnt]=ed,nxt[cnt]=head[bg],head[bg]=cnt;
}
int main(){
// freopen("data.txt","r",stdin);
n=rd();int x,y;
for(register int i=1;i<n;i++) {
x=rd(),y=rd();
add(x,y),add(y,x);
}
for(register int i=1;i<=n;i++) val[i]=rd();
for(register int i=1;i<=n;i++){
LL sum=0,now=0;
for(register int j=head[i];j;j=nxt[j]){
int u=to[j];
if(val[u]*now>mx) mx=val[u]*now;
if(val[u]>now) now=val[u];
ans+=val[u]*sum;ans%=mod;
sum+=val[u];sum%=mod;
}
}
printf("%lld %lld",mx,ans*2%mod);
return 0;
}