P1351 联合权值

因为距离为2，所以枚举中间点即可。

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define maxn 200010

using namespace std;

struct node
{
    int ed,nxt;
};
node edge[maxn<<1];
int n,m,first[maxn],cnt;
long long w[maxn],d[maxn];
const long long mod=10007;
long long ans=0,sum=0;

inline void add_edge(int s,int e)
{
    ++cnt;
    edge[cnt].ed=e;
    edge[cnt].nxt=first[s];
    first[s]=cnt;
    return;
}

int main()
{
    scanf("%d",&n);
    for(register int i=1;i<=n-1;++i)
    {
        int s,e;
        scanf("%d%d",&s,&e);
        add_edge(s,e);
        add_edge(e,s);
        d[s]++;
        d[e]++;
    }
    for(register int i=1;i<=n;++i)
        scanf("%lld",&w[i]);
    for(register int i=1;i<=n;++i)
    {
        if(d[i]==1) continue;
        long long max_1=0,max_2=0,sum1=0,sum2=0;
        for(register int j=first[i];j;j=edge[j].nxt)
        {
            //cout<<max_1<<' '<<max_2<<endl;
            int e=edge[j].ed;
            if(w[e]>max_1) max_2=max_1,max_1=w[e];
            else if(w[e]>max_2) max_2=w[e];
            sum1=(sum1+w[e])%mod;
            sum2=(sum2+w[e]*w[e])%mod;
            //cout<<max_1<<' '<<max_2<<endl;
        }
        sum1=sum1*sum1%mod;
        sum=(sum+sum1+mod-sum2)%mod;
        ans=max(ans,max_1*max_2);
        //cout<<max_1<<' '<<max_2<<endl;
    }
    printf("%lld %lld\n",ans,sum); 
    return 0;
}

标签：node,include,max,sum2,sum1,联合,权值,P1351,mod
来源： https://www.cnblogs.com/Hoyoak/p/11835774.html