牛客 - 点对最大值

传送门：牛客 - 点对最大值

思路分析

显然可以进行DP，$dp[u]$表示以$u$为起点向下权值最大的链

AC代码

#include <bits/stdc++.h>
#define ls x<<1
#define rs x<<1|1
#define fi first
#define se second
#define ll long long
#define pb push_back
#define mp make_pair
#define fun function
#define sz(x) (x).size()
#define lowbit(x) (x)&(-x)
#define all(x) (x).begin(),(x).end()
#define mem(a,b) memset(a,b,sizeof(a))

namespace FastIO {
#define BUF_SIZE 100000
#define OUT_SIZE 100000
    bool IOerror=0;
    inline char nc() {
        static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE;
        if(p1==pend) {
            p1=buf;
            pend=buf+fread(buf,1,BUF_SIZE,stdin);
            if(pend==p1) {
                IOerror=1;
                return -1;
            }
        }
        return *p1++;
    }
    inline bool blank(char ch) {
        return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';
    }
    template<class T> inline bool read(T &x) {
        bool sign=0;
        char ch=nc();
        x=0;
        for(; blank(ch); ch=nc());
        if(IOerror)return false;
        if(ch=='-')sign=1,ch=nc();
        for(; ch>='0'&&ch<='9'; ch=nc())x=x*10+ch-'0';
        if(sign)x=-x;
        return true;
    }
    template<class T,class... U>bool read(T& h,U&... t) {
        return read(h)&&read(t...);
    }
#undef OUT_SIZE
#undef BUF_SIZE
};
using namespace std;
using namespace FastIO;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int INF = 0x3f3f3f3f;
const int N = 1e6+10;

vector<pair<int,int>>e[N];
int a[N],dp[N],ans; 

void dfs(int u){
    dp[u]=a[u];
    for(auto x:e[u]){
        int v=x.fi;
        int w=x.se;
        dfs(v);
        ans=max(ans,dp[v]+dp[u]+w);
        dp[u]=max(dp[u],dp[v]+w);
    }
}

int main() {

#ifdef xiaofan
    freopen("1.in","r",stdin);
    freopen("1.out","w",stdout);
#endif

    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        for(int i=2;i<=n;i++){
            int x,y;
            cin>>x>>y;
            e[x].pb(mp(i,y));
        }
        ans=-INF;
        for(int i=1;i<=n;i++) cin>>a[i];
        dfs(1);
        cout<<ans<<endl;
        for(int i=1;i<=n;i++) e[i].clear();
    }




    return 0;
}