来源:POJ1330
使用dfs序来寻找子树的信息,随后就可以完成这个答案了
代码:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
using namespace std;
const int MAXN = 10000+10;
int n;
vector<int>tree[MAXN];
bool vis[MAXN];
int in[MAXN],dep[MAXN<<1],has[MAXN<<1];
int dp[MAXN<<1][20];
int id;
void dfs(int rt,int depth){
dep[++id] = depth;
has[id] = rt;
if(!in[rt]) in[rt] = id;
int nxt;
for(int i=0;i<tree[rt].size();i++){
nxt = tree[rt][i];
dfs(nxt,depth+1);
has[++id] = rt;
dep[id] = depth;
}
return ;
}
void stpre(){
//cout<<"id = "<<id<<endl;
for(int i=1;i<=id;i++) dp[i][0] = i;
for(int l=1;(1<<l)<=id;l++){
for(int i=1;i<=id;i++){
if(i+(1<<l)-1<=id){
int xx = dp[i][l-1];
int yy = dp[i+(1<<(l-1))][l-1];
if(dep[xx]>dep[yy]) dp[i][l] = yy;
else dp[i][l] = xx;
}
}
}
}
int query(int l,int r){
int k=(int)(log(double(r-l+1))/log(2.0));
int xx = dp[l][k];
int yy = dp[r-(1<<k)+1][k];
//cout<<"id: "<<xx<<" "<<yy<<endl;
if(dep[xx]>dep[yy]) return has[yy];
else return has[xx];
}
int main(){
int T;
int x,y;
scanf("%d",&T);
while(T--){
memset(vis,false,sizeof(vis));
scanf("%d",&n);
for(int i=1;i<=n;i++) tree[i].clear();
for(int i=1;i<n;i++){
scanf("%d%d",&x,&y);
tree[x].push_back(y);
vis[y] = true;
}
int root = -1;
for(int i=1;i<=n;i++){
if(!vis[i]){
root = i;
break;
}
}
id = 0;
memset(dep,0,sizeof(dep));
memset(in,0,sizeof(in));
memset(has,0,sizeof(has));
memset(dp,0,sizeof(dp));
dfs(root,1);
//cout<<"---------"<<endl;
//for(int i=1;i<=id;i++){
// cout<<i<<" "<<has[i]<<" "<<dep[i]<<endl;
//}
//cout<<"----------"<<endl;
//cout<<"2*n-1 = "<<2*n-1<<" id = "<<id<<endl;
//for(int i=1;i<=n;i++) cout<<i<<" "<<in[i]<<endl;
stpre();
scanf("%d%d",&x,&y);
if(in[x]>in[y]) cout<<query(in[y],in[x])<<endl;
else cout<<query(in[x],in[y])<<endl;
}
return 0;
}