日常猝死。
A：f[i]表示子树内包含根且可以继续向上延伸的路径的最大价值，统计答案考虑合并两条路径即可。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using namespace std; #define ll long long #define N 300010 #define inf 10000000000000000ll char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int n,p[N],t; ll f[N][2],a[N],ans; struct data{int to,nxt,len; }edge[N<<1]; void addedge(int x,int y,int z){t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].len=z,p[x]=t;} void dfs(int k,int from) { f[k][1]=f[k][0]=a[k]; ll mx=-inf,mx2=-inf; for (int i=p[k];i;i=edge[i].nxt) if (edge[i].to!=from) { dfs(edge[i].to,k); ll x=f[edge[i].to][1]-edge[i].len; if (x>mx) mx2=mx,mx=x; else if (x>mx2) mx2=x; } f[k][1]=max(mx+a[k],a[k]); f[k][0]=max(f[k][0],f[k][1]); f[k][0]=max(f[k][0],a[k]+mx+mx2); } int main() { #ifndef ONLINE_JUDGE freopen("a.in","r",stdin); freopen("a.out","w",stdout); #endif n=read(); for (int i=1;i<=n;i++) a[i]=read(); for (int i=1;i<n;i++) { int x=read(),y=read(),z=read(); addedge(x,y,z),addedge(y,x,z); } dfs(1,1); for (int i=1;i<=n;i++) ans=max(ans,f[i][0]),ans=max(ans,f[i][1]); cout<<ans; return 0; }