2016夏季联系——LCA+dfs_order

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;}