【洛谷P1351】联合权值

我们枚举中间点，当连的点数不小于2时进行处理

最大值好搞

求和：设中间点 i 所连所有点权之和为sum

则对于每个中间点i的联合权值之和为：

w[j]*(sum-w[j])之和

 #include<cstdio>
 #include<cstring>
 using namespace std;
 const int p=,N=,M=;
 int head[M],next[M],to[M],du[N],a[N],size;
 int w[N],n,sum,ss,m1,m2,ans1,ans2;
 void uni(int x,int y){
     size++;
     du[x]++;
     next[size]=head[x];
     head[x]=size;
     to[size]=y;
 }
 int main(){
     int x,y,q;
     size=ans1=;
     scanf("%d",&n);
     for (int i=;i<n;i++){
         scanf("%d %d",&x,&y);
         uni(x,y);
         uni(y,x);
     }
     for (int i=;i<=n;i++)
         scanf("%d",&w[i]);
     for (int x=;x<=n;x++){
         if (du[x]<)
             continue;
         m1=m2=sum=q=ss=;
         for (int e=head[x];e;e=next[e]){
             int y=to[e];
             sum=(sum+w[y])%p;
             a[++q]=y;
             if (w[y]>m1){
                 m2=m1;
                 m1=w[y];
             }
             else if (w[y]>m2)
                 m2=w[y];
         }
         if (m1*m2>ans1)
             ans1=m1*m2;
         for (int j=;j<=q;j++)
             ans2=(ans2+w[a[j]]*(sum-w[a[j]]+p)%p)%p;
     }
     printf("%d %d",ans1,ans2);
     return ;
 }

STD