【題解】洛谷P1351[NOIP2014]聯合權值 亂搞

題目鏈接

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處理出每個點相鄰點的和，然後直接搞一搞

#include<cstdio>
#include<vector>
#include<algorithm>
using namespace std;
const int N=2e5+10,mod=10007;
inline int read()
{
	int x=0,f=0;char ch=getchar();
	while(ch<'0'||ch>'9')f|=ch=='-',ch=getchar();
	while(ch>='0'&&ch<='9')x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	if(f)x=-x;return x;
}
int n,sum[N],w[N],m1,m2,maxn,ans;
vector<int>g[N];
int main()
{
	//freopen("in.txt","r",stdin);
	n=read();
	for(int i=1,u,v;i<n;i++)
	    u=read(),v=read(),g[u].push_back(v),g[v].push_back(u);
	for(int i=1;i<=n;i++)
	    w[i]=read();
	for(int i=1;i<=n;i++){
		m1=m2=0;
		for(int j=0;j<g[i].size();j++)
		{
			sum[i]=(sum[i]+w[g[i][j]])%mod;
			if(w[g[i][j]]>m1)m2=m1,m1=w[g[i][j]];
			else if(w[g[i][j]]>m2)m2=w[g[i][j]];
		}
		maxn=max(maxn,m1*m2);
    }
    printf("%d ",maxn);
	for(int i=1;i<=n;i++)
	    for(int j=0;j<g[i].size();j++)
	        ans=(ans+(sum[i]-w[g[i][j]]+mod)%mod*w[g[i][j]]%mod)%mod;
	printf("%d\n",ans);
	return 0;
}

總結

水題