hdu5266 pog loves szh III 【LCA】【倍增】

Pog and Szh are playing games. Firstly Pog draw a tree on the paper. Here we define 1 as the root of the tree.Then Szh choose some nodes from the tree. He wants Pog helps to find the least common ancestor (LCA) of these node.The question is too difficult for Pog.So he decided to simplify the problems.The nodes picked are consecutive numbers from lili to riri ([li,ri])([li,ri]). 

Hint : You should be careful about stack overflow !

InputSeveral groups of data (no more than 3 groups,n≥10000n≥10000 or Q≥10000Q≥10000). 

The following line contains ans integers,n(2≤n≤300000)n(2≤n≤300000). 

AT The following n−1n−1 line, two integers are bibi and cici at every line, it shows an edge connecting bibi and cici. 

The following line contains ans integers,Q(Q≤300000)Q(Q≤300000). 

AT The following QQ line contains two integers li and ri(1≤li≤ri≤n1≤li≤ri≤n).OutputFor each case,output QQ integers means the LCA of [li,ri][li,ri].Sample Input

5
1 2
1 3
3 4
4 5
5
1 2
2 3
3 4
3 5
1 5

Sample Output

1
1
3
3
1



        
  

Hint

Be careful about stack overflow.
        

用BFS 预处理不会爆栈  注意看n的范围 看到前面的1e5就以为是1e5了

f[i][j]表示节点i的第2^j个祖先 

#include <iostream>
#include <algorithm>
#include <stdlib.h>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <stdio.h>
#include <queue>
#include <stack>
#define inf 0x3f3f3f3f
using namespace std;

int n, q, ecnt;
const int maxn = 300005;
struct edge{
    int v, next;
}e[maxn << 1];
int dep[maxn], f[20][maxn], head[maxn];


void bfs(int rt)
{
    queue<int> q;
    q.push(rt);
    f[0][rt] = rt;
    dep[rt] = 0;
    while(!q.empty()){
        int tmp = q.front();
        q.pop();
        for(int i = 1; i < 20; i++){
            f[i][tmp] = f[i - 1][f[i - 1][tmp]];
        }
        for(int i = head[tmp]; i != -1; i = e[i].next){
            int v = e[i].v;
            if(v == f[0][tmp]) continue;
            dep[v] = dep[tmp] + 1;
            f[0][v] = tmp;
            q.push(v);
        }
    }
}

int LCA(int u, int v)
{
    if(dep[u] > dep[v]) swap(u, v);
    int hu = dep[u], hv = dep[v];
    int tu = u, tv = v;
    for(int det = hv - hu, i = 0; det; det >>= 1, i++){
        if(det & 1){
            tv = f[i][tv];
        }
    }
    if(tu == tv){
        return tu;
    }
    for(int i = 19; i >= 0; i--){
        if(f[i][tu] == f[i][tv]){
            continue;
        }
        tu = f[i][tu];
        tv = f[i][tv];
    }
    return f[0][tu];

}

void init()
{
    memset(head, -1, sizeof(head));
    ecnt = 0;
}

void adde(int u, int v)
{
    e[ecnt].v = v;
    e[ecnt].next = head[u];
    head[u] = ecnt++;
}

int dp[maxn][20];
int main()
{
    while(scanf("%d", &n) != EOF){
        init();
        for(int i = 1; i < n; i++){
            int x, y;
            scanf("%d%d", &x, &y);
            adde(x, y);
            adde(y, x);
        }
        bfs(1);
        for(int i = 1; i <= n; i++){
            dp[i][0] = i;
        }
        for(int j = 1; j < 20; j++){
            for(int i = 1; i + (1 << j) - 1 <= n; i++){
                dp[i][j] = LCA(dp[i][j - 1], dp[i + (1 << (j - 1))][j - 1]);
            }
        }

        scanf("%d", &q);
        while(q--){
            int l, r;
            scanf("%d%d", &l, &r);
            int k = (int)log2(r - l + 1);
            cout<<LCA(dp[l][k], dp[r - (1 << k) + 1][k])<<endl;
        }
    }
    return 0;
}