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数据范围
1≤n≤10^5,
每个节点的评分的绝对值均不超过 10^6。
输入样例：
5 1 -2 -3 4 5 4 2 3 1 1 2 2 5
输出样例：
8
分析：这道题是要我们在树中求一个最大连通块，我们可以定义f[i]为以i为根的子树中最大连通块的值
，这样结果就是f[1~n]中的最大值，树形DP过程比较简单，一开始令f[i]=w[i]，也就是令这个连通块只包含自己这一个点，然后只要以子节点为根的子树中最大连通块的值大于0，就加上，按照这样进行dp就可以求出答案，细节参照代码：
#include<cstdio> #include<iostream> #include<cstring> #include<vector>
#include<algorithm> #include<map> #include<cmath> #include<queue> using
namespace std; #define int long long const int N=1e5+10; int
w[N*2],f[N],e[N*2],ne[N*2],h[N],idx;//f[i]表示以i为根的连通块中的最大值 void add(int x,int y)
{ e[idx]=y; ne[idx]=h[x]; h[x]=idx++; } int dp(int x,int father) { f[x]=w[x];
for(int i=h[x];i!=-1;i=ne[i]) { int j=e[i]; if(j==father) continue;
f[x]+=max(dp(j,x),(int)0); } return f[x]; } signed main() { int n; cin>>n;
for(int i=1;i<=n;i++) scanf("%lld",&w[i]); memset(h,-1,sizeof h); for(int
i=1;i<n;i++) { int u,v; scanf("%lld%lld",&u,&v); add(u,v);add(v,u); } dp(1,0);
int ans=-0x3f3f3f3f; for(int i=1;i<=n;i++) ans=max(ans,f[i]);
printf("%lld",ans); return 0; }
