uval 1752 tree 2

找到树的直径，然后分别从直径两个端点建立倍增数组实现查找。

#include <bits/stdc++.h>
#define  pb push_back
#define  lson l,m,rt<<1
#define  rson m+1,r,rt<<1|1
typedef  long long LL;
using namespace std;
const int maxn = 20010;
vector<int>G[maxn];
int dp[maxn][20][2];
int pre[maxn][20][2];
int dis[maxn][20][2];
bool vis[maxn];
int l,r,n;
void bfs()
{
    queue<pair<int,int> >Q;
    Q.push( make_pair(1,0) );
    memset(vis,0,sizeof(vis));
    vis[1] = 1;
    int num = 0,ans;
    while(!Q.empty())
    {
        int a = Q.front().first;
        int b = Q.front().second;
       if(b>num){
            num = b;
            ans = a;
        }
        Q.pop();
        for(int i=0;i<G[a].size();i++)
        {
            int v = G[a][i];
            if(!vis[v])
            {
                vis[v] = 1;
                Q.push(make_pair(v,b+1));
            }
        }
    }
    l = ans;
    memset(vis,0,sizeof(vis));
    vis[l] = 1;
    Q.push(make_pair(l,0));
    num = 0;
    while(!Q.empty())
    {
        int a = Q.front().first;
        int b = Q.front().second;
        if(b>num){
            num = b;
            ans = a;
        }
        Q.pop();
        for(int i=0;i<G[a].size();i++)
        {
            int v = G[a][i];
            if(!vis[v])
            {
                vis[v] = 1;
                Q.push(make_pair(v,b+1));
            }
        }
    }
    r = ans;
}
void dfs(int x,int fa,int de,int flag)
{
    vis[x] = 1;
    pre[x][0][flag] = fa;
    for(int i=0;i<G[x].size();i++)
    if(G[x][i]!=fa)
    {
        dis[ G[x][i] ][0][flag] = 1;
        dfs(G[x][i],x,de+1,flag);
    }
}
void init()
{
    memset(pre,-1,sizeof(pre));
    memset(dis,0,sizeof(dis));
    memset(vis,0,sizeof(vis));
    dfs(l,l,1,0);
     for(int k=1; k<20; ++k)
        for(int i=1; i<=n; ++i)
            if(pre[i][k-1][0]!=-1)
            {
                pre[i][k][0] = pre[ pre[i][k-1][0] ][k-1][0];
                dis[i][k][0] = dis[i][k-1][0] + dis[ pre[i][k-1][0] ][k-1][0];
            }
    memset(vis,0,sizeof(vis));
    dfs(r,r,1,1);
    for(int k=1; k<20; ++k)
        for(int i=1; i<=n; ++i)
            if(pre[i][k-1][1]!=-1)
            {
                pre[i][k][1] = pre[ pre[i][k-1][1] ][k-1][1];
                dis[i][k][1] = dis[i][k-1][1] + dis[ pre[i][k-1][1] ][k-1][1];
            }
}
void solve(int x,int d)
{
    if(d==0)
    {
        printf("%d\n",x);
        return ;
    }
    int ansd = d,y=x;
    while(1)
    {
        int j = -1;
        for(int k=0;k<20;k++)
        if(dis[y][k][0]>=ansd)
        {
            j = k;
            break;
        }
        if(j==-1) break;
        if(dis[y][j][0]==ansd)
        {
            printf("%d\n",pre[y][j][0]);
            return ;
        }
        else {
            ansd -= dis[y][j-1][0];
            y = pre[y][j-1][0];
        }
    }
      ansd = d,y=x;
     while(1)
    {
        int j = -1;
        for(int k=0;k<20;k++)
        if(dis[y][k][1]>=ansd)
        {
            j = k;
            break;
        }
        if(j==-1) break;
        if(dis[y][j][1]==ansd)
        {
            printf("%d\n",pre[y][j][1]);
            return ;
        }
        else {
            ansd -= dis[y][j-1][1];
            y = pre[y][j-1][1];
        }
    }
    printf("0\n");
}
int main()
{
    int Q,x,y,val,d;
    while(scanf("%d%d",&n,&Q)!=EOF)
    {
        for(int i=0;i<=n;i++) G[i].clear();
        for(int i=1;i<n;i++)
        {
            scanf("%d%d",&x,&y);
            G[x].pb(y);
            G[y].pb(x);
        }
        bfs();
        init();
        while(Q--)
        {
            scanf("%d%d",&val,&d);
            solve(val,d);
        }
    }
    return 0;
}