HDU 4008 Parent and son LCA+树形dp

题意：

给定case数

给定n个点的树，m个询问

以下n-1行给出树边

m个询问 x y

问：以x为根。y子树下 y的最小点标的儿子节点 和子孙节点

思路：

用son[u][0] 表示u的最小儿子 son[u][2] 表示u的次小儿子

son[u][1] 表示u的最小子孙

若lca(x,y)  !=y  则就是上述的答案

若lca(x,y) == y

1、y != 1 那么最小儿子就是除了x外的节点，且此时father[y] 也是y的儿子节点， 而最小的子孙节点就是1

2、y==1 那么特殊处理一下。用map维护一下除了u节点下的子树。1的最小子孙节点

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define inf 1000000
#define N 100050
struct Edge{
    int from, to, dis, nex;
}edge[5*N];
int head[N],edgenum,dis[N],fa[N][20],dep[N];  //fa[i][x] 是i的第2^x个父亲（假设超过范围就是根）
void add(int u,int v,int w){
    Edge E={u,v,w,head[u]};
    edge[edgenum] = E;
    head[u]=edgenum++;
}
void bfs(int root){
    queue<int> q;
    fa[root][0]=root;dep[root]=0;dis[root]=0;
    q.push(root);
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int i=1;i<20;i++)fa[u][i]=fa[fa[u][i-1]][i-1];
        for(int i=head[u]; ~i;i=edge[i].nex){
            int v=edge[i].to;if(v==fa[u][0])continue;
            dep[v]=dep[u]+1;dis[v]=dis[u]+edge[i].dis;fa[v][0]=u;
            q.push(v);
        }
    }
}
int move(int x,int y){//把x移动到y以下
    int D = dep[x] - dep[y]-1;
    while(D){
        int z = (int)(log10(1.0*D)/log10(2.0));
        x = fa[x][z];
        D-=1<<z;
    }
    return x;
}
int Lca(int x,int y){
    if(dep[x]<dep[y])swap(x,y);
    for(int i=0;i<20;i++)if((dep[x]-dep[y])&(1<<i))x=fa[x][i];
    if(x==y)return x;
    for(int i=19;i>=0;i--)if(fa[x][i]!=fa[y][i])x=fa[x][i],y=fa[y][i];
    return fa[x][0];
}
void init(){memset(head, -1, sizeof head); edgenum = 0;}
int son[N][3], Fa[N];
//0是最小 1是子孙最小 2是次小
void dfs(int u,int  father){
	Fa[u] = father;
    son[u][0] = son[u][1] = son[u][2] = inf;
    for(int i = head[u]; ~i; i = edge[i].nex){
        int v = edge[i].to; if(v==father)continue;
        dfs(v,u);
        if(v<son[u][0])
			son[u][2] = son[u][0], son[u][0] = v;
		else son[u][2] = min(son[u][2], v);
        son[u][1] = min(son[u][1], son[v][1]);
    }
    son[u][1] = min(son[u][1], son[u][0]);
}
map<int,int>mp;
bool work(int x,int y){
    int lca = Lca(x,y), son1, son2;
        if(lca==y) {
	        x = move(x,y);
	        son1 = son[y][0];
	        if(son1 == x) son1 = son[y][2];
	        son1 = min(son1, Fa[y]);
	        if(y==1) {
	        	son2 = mp[x];
			}
			else son2 = 1;
  	    } else {
  	    	son1 = son[y][0];
  	    	son2 = son[y][1];
		}
	if(son1==inf||son2==inf)return false;
    printf("%d %d\n",son1,son2);
    return true;
}
int n;
int main(){
    int i,j,u,v,T,que;scanf("%d",&T);
    while(T--){
        init();
        scanf("%d %d",&n,&que);
        for(i = 1; i < n; i++) {
            scanf("%d %d",&u,&v);
            add(u,v,1);
            add(v,u,1);
        }
        bfs(1);
        dfs(1,1);
        Fa[1] = inf;
        mp.clear();
        int fir = inf, sec = inf;
        for(i = head[1]; ~i; i = edge[i].nex){
			v = edge[i].to;
			mp[v] = inf;
			if(v<fir)sec = fir, fir = v;
			else sec = min(sec, v);
		}

		for(i = head[1]; ~i; i = edge[i].nex){
			v = edge[i].to;
			if(v!=fir)mp[v] = fir;
			else mp[v] = sec;
		}
		fir = inf; sec = inf;
	    for(i = head[1]; ~i; i = edge[i].nex){
			v = edge[i].to;
			if(son[v][1]<fir)sec = fir, fir = son[v][1];
			else sec = min(sec, son[v][1]);
		}
	    for(i = head[1]; ~i; i = edge[i].nex){
			v = edge[i].to;
			if(son[v][1]!=fir)
				mp[v] = min(mp[v], fir);
			else
				mp[v] = min(mp[v], sec);
  		}
        while(que--){
            scanf("%d %d",&u,&v);
            if(work(u,v)==false)
                puts("no answers!");
        }
        puts("");
    }
    return 0;
}