【Codeforces418D】Big Problems for Organizers

先求出树的直径。
预处理出直径,用 rmq 维护处直径上某一个区间点的子树内到直径两端点距离的最大值。
询问时,先找到 x,y 两点所在的根,然后答案只能是一下两种:
1、直径两端点的答案
2、 x 所在树到路径中点所在直径上的点的树中距x的最大距离或者另一半的同样情况。

#include <bits/stdc++.h>
#define gc getchar()
#define ll long long
#define mid (l+r>>1)
#define N 100009
using namespace std;
int n,Q,first[N],number,root[N],fa[N],dis[N],f[N][20],g[N][20],bit[20],lg[N],X,Y;
struct edge
{
    int to,next;
    void add(int x,int y)
    {
        to=y,next=first[x],first[x]=number;
    }
}e[N<<1];
int read()
{
    int x=1;
    char ch;
    while (ch=gc,ch<'0'||ch>'9') if (ch=='-') x=-1;
    int s=ch-'0';
    while (ch=gc,ch>='0'&&ch<='9') s=s*10+ch-'0';
    return s*x;
}
int bfs(int x,int from)
{
    int ret=0;
    queue<int> q;
    while (!q.empty()) q.pop();
    dis[x]=0,root[x]=from,fa[x]=from;
    q.push(x);
    while (!q.empty())
    {
        int now=q.front();
        q.pop();
        if (dis[ret]<dis[now]) ret=now;
        for (int i=first[now];i;i=e[i].next)
            if (e[i].to!=fa[now])
            {
                dis[e[i].to]=dis[now]+1,root[e[i].to]=from,fa[e[i].to]=now;
                q.push(e[i].to);
            }
    }
    return ret;
}
int rmq(int f[][20],int x,int y)
{
    if (x>y) return 0;
    int k=lg[y-x+1];
    return max(f[x][k],f[y-bit[k]+1][k]);
}
void init()
{
    //ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);
    bit[0]=1;
    for (int i=1;i<20;i++) bit[i]=bit[i-1]<<1;
    lg[0]=-1;
    for (int i=1;i<N;i++) lg[i]=lg[i>>1]+1;
    n=read();
    for (int i=1;i<n;i++)
    {
        int x=read(),y=read();
        e[++number].add(x,y),e[++number].add(y,x);
    }
    int x=bfs(1,0);
    int y=bfs(x,0);
    for (int i=y,j=0;i;j=i,i=fa[i])
    {
        int Max=0;
        root[i]=i;
        for (int k=first[i];k;k=e[k].next)
            if (e[k].to!=fa[i]&&e[k].to!=j)
                Max=max(Max,dis[bfs(e[k].to,i)]+1);
        f[dis[i]][0]=Max+dis[i];
        g[dis[i]][0]=Max+dis[y]-dis[i];
    }
    for (int i=1;i<20;i++)
        for (int j=1;j+bit[i]-1<=dis[y];j++)
        {
            f[j][i]=max(f[j][i-1],f[j+bit[i-1]][i-1]);
            g[j][i]=max(g[j][i-1],g[j+bit[i-1]][i-1]);
        }
    X=x,Y=y;
}
void work()
{
    Q=read();
    while (Q--)
    {
        int x=read(),y=read();
        int l=dis[root[x]],r=dis[root[y]];
        if (l<r) swap(l,r),swap(x,y);
        int dx=0,dy=0;
        if (x!=root[x]) l+=dis[x]+1,dx=dis[x]+1,x=root[x];
        if (y!=root[y]) r-=dis[y]+1,dy=dis[y]+1,y=root[y];
        int ret=min(dx+dis[Y]-dis[x],dy+dis[Y]-dis[y]);
        ret=max(ret,min(dx+dis[x],dy+dis[y]));
        ret=max(ret,rmq(f,dis[y]+1,min(mid,dis[x]))-r);
        ret=max(ret,rmq(g,max(mid+1,dis[y]),dis[x]-1)-(dis[Y]-l));
        printf("%d\n",ret);
    }
}
int main()
{
    init();
    work();
    return 0;
}
### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?
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