学军信友队趣味网络邀请赛 A-B-D 思维+树形DP/直径+数论

题目链接：http://115.236.49.52:83/contest/1351


题解：假设n是奇数。n如果是偶数，翻转90度就可以了。

B:
#include 
using namespace std;
#define LL long long
 
vector<vector<int> > v(50005);
int a[50005];
LL f[50005][1005];
LL ans=0;
void DFS(int u, int fa){
 
    f[u][a[u]]=max(f[u][a[u]], 0ll);
    for(auto x: v[u]){
        if(x!=fa){
            DFS(x, u);
            LL mx1=0, mx2=0;
            for(int i=1; i<=1000; i++){
 
                if(f[x][i]!=-1) f[x][i]+=1;
                mx1=max(mx1, f[u][i]); mx2=max(mx2, f[x][i]);
                if(f[x][i]!=-1){
                    ans=max(ans, (f[x][i]+mx1)*i);
                }
                if(f[u][i]!=-1){
                    ans=max(ans, (f[u][i]+mx2)*i);
                }
                if(f[x][i]!=-1) f[u][i]=max(f[u][i], f[x][i]);
            }
        }
    }
}
 
int main() {
 
    memset(f, -1, sizeof(f));
    int n, x, y; scanf("%d", &n);
    for(int i=1; i<=n; i++){
        scanf("%d", &a[i]);
    }
    for(int i=1; i<n; i++){
        scanf("%d%d", &x, &y);
        v[x].push_back(y);
        v[y].push_back(x);
    }
    DFS(1, 0);
    printf("%lld\n", ans);
 
    return 0;
}
 
 
 
 
#include 
using namespace std;
#define LL long long
const int maxn=5e4+10;
 
struct E{
    int to, w;
    E(int a, int b){
        to=a; w=b;
    }
};
vector<E> e[maxn];
int k, d, pos1, pos2;
int vis[maxn], a[maxn];
 
void dfs(int u, int fa, int s){
    vis[u]=1;
    if(s>=d){
        d=s, k=u;
    }
    for(int i=0;i<e[u].size();i++){
        int v=e[u][i].to, w=e[u][i].w;
        if(v!=fa){
            dfs(v, u, s+w);
        }
    }
}
 
LL ans=0;
void dfs(int u, int fa, LL x, LL s){
    if(a[u]>=a[x]){
        ans=max(ans, s*a[u]);
    }
    for(auto to: e[u]){
        if(to.to!=fa){
            dfs(to.to, u, x, s+to.w);
        }
    }
}
 
int main()
{
    int n, u, v, w;
    scanf("%d", &n);
    for(int i=1; i<=n; i++){
        scanf("%d", &a[i]);
    }
    for(int i=1; i<n; i++){
        scanf("%d%d",&u,&v);w=1;
        e[u].push_back(E(v, w));
        e[v].push_back(E(u, w));
    }
    pos1=pos2=k=0;
    dfs(1, 0, 0); pos1=k;
    dfs(k, 0, 0); pos2=k;
    dfs(pos1, 0, pos1, 0);
    dfs(pos2, 0, pos2, 0);
    printf("%lld\n", ans);
 
    return 0;
}