本文主要是介绍Tourism on Mars URAL - 2109 （LCA，线段树），希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
LCA博客 https://blog.csdn.net/sdut_jk17_zhangming/article/details/84592285
Few people know, but a long time ago a developed state existed on Mars. It consisted of n cities, numbered by integers from 1 to n, the capital had the number 1. Some pairs of cities were connected by a road. The residents of the state were very prudent, therefore, between any two cities, there was exactly one path (possibly consisting of several roads).
Due to the fact that the state was developed, its residents loved traveling. Tourist route on Mars was described by two numbers L and R. This meant that the tourist started the route in the city L, then went to the city L + 1 (without going into the cities, that did not lie on the path between L and L + 1), then went to the city L + 2, and so on. The last city on the route was the city R. A city that was the closest to the capital among all cities visited on this route (if to count a distance between the cities by the roads) was considered the main attraction of the route.
Knowing the map of the Martian state and all the tourist routes, find for each route its main attraction.
Input
The first line contains an integer n that is the number of cities (1 ≤ n ≤ 2 · 105).
The following n − 1 lines describe the roads. Each road is described by two numbers of cities that are connected by it (1 ≤ v i, u i ≤ n; v i ≠ u i).
The ( n + 1)-th line contains an integer q that is the number of tourist routes (0 ≤ q ≤ 10 6).
Then q lines describe the routes themselves. Each route is described by a pair of integers L i, R i (1 ≤ L i ≤ R i ≤ n).
Output
Output q integers, one per line — for each route the number of its main attraction. These numbers should be output in the same order in which the respective routes were described.
Example
| input | output |
|---|---|
7 1 2 1 3 2 4 2 5 2 6 3 7 3 4 6 3 4 5 7 | 2 1 1 |
7 1 3 3 5 5 6 7 5 1 4 2 4 3 4 5 5 6 6 7 | 1 5 5 |
Notes
This problem has a big input and output data size and a strict Time Limit. If you write your solution in C++ we recommend you to use Visual C++ 2013 compiler.
#include<cstdio>
#include<vector>
#include <bits/stdc++.h>
using namespace std;
struct node
{int v,next;
}e[5000005];
int tr[2000006];
int n;
int cnt;
int head[200005];
int dep[200005];
int f[200005][30];
void add(int u,int v)
{e[cnt] = node{v,head[u]};head[u] = cnt++;
}
void dfs(int x,int p,int d)
{f[x][0] = p;dep[x] = d;for(int i = head[x];i!=-1;i=e[i].next){int to = e[i].v;if(to!=p){dfs(to,x,d+1);}}
}
void init()
{dfs(1,0,0);for(int i=1;i<=21;i++){for(int j = 1;j<=n;j++){f[j][i] = f[f[j][i-1]][i-1];}}
}
int lca(int x,int y)
{if(dep[x] < dep[y]) swap(x,y);for(int i=21;i>=0;i--){if(dep[x]>dep[y]&&dep[f[x][i]] >= dep[y]&&f[x][i] != 0){x = f[x][i];}}if(x == y) return x;for(int i=21;i>=0;i--){if(f[x][i] != f[y][i]){x = f[x][i];y = f[y][i];}}return f[x][0];
}
void build(int rt,int l,int r)
{if(l==r){tr[rt] = lca(l,l+1);return ;}int mid = (l+r)>>1;build(rt<<1,l,mid);build(rt<<1|1,mid+1,r);if(dep[tr[rt<<1]] < dep[tr[rt<<1|1]]){tr[rt] = tr[rt<<1];}else tr[rt] = tr[rt<<1|1];
}
int minn,ans;
void query(int rt,int l,int r,int L,int R)
{if(L <= l&& R >= r){if(dep[tr[rt]] < minn){minn = dep[tr[rt]];ans = tr[rt];}return ;}int mid = (l+r)>>1;if(L <= mid) query(rt<<1,l,mid,L,R);if(R > mid) query(rt<<1|1,mid+1,r,L,R);
}
int main()
{scanf("%d",&n);int i,u,v;cnt = 0;for(i=1;i<=n;i++){head[i] = -1;}for(i=1;i<n;i++){scanf("%d%d",&u,&v);add(u,v);add(v,u);}if(n==1){scanf("%d",&n);while(n--){scanf("%d%d",&u,&v);printf("1\n");}return 0;}memset(f,0,sizeof(f));init();build(1,1,n-1);int m;scanf("%d",&m);int a,b;while(m--){scanf("%d%d",&a,&b);if(a == b){printf("%d\n",a);continue;}minn = 0x3f3f3f3f;query(1,1,n-1,a,b-1);printf("%d\n",ans);}return 0;
}
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