【spoj】【QTREE2 - Query on a tree II】

题目大意

给出一棵树，每条边有边权，询问两点之间的距离，以及从起点到终点第k个点是那个点。

解题思路

观察可知，整个询问是静态的，可以使用倍增算法解决。

code

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LF double
#define LL long long
#define Min(a,b) ((a<b)?a:b)
#define Max(a,b) ((a>b)?a:b)
#define Fo(i,j,k) for(int i=j;i<=k;i++)
#define Fd(i,j,k) for(int i=j;i>=k;i--)
using namespace std;
int const MxN=1e4;
int N,M,Edge,Mx,Two[20],Begin[MxN+9],Dep[MxN+9],Fa[MxN+9][17],Dis[MxN+9][17],To[MxN*2+9],Len[MxN*2+9],Next[MxN*2+9];
void Insert(int U,int V,int W){
To[++Edge]=V;
Len[Edge]=W;
Next[Edge]=Begin[U];
Begin[U]=Edge;
}
void Dfs(int Now,int Pre){
Dep[Now]=Dep[Pre]+1;
for(int i=Begin[Now];i;i=Next[i])if(To[i]!=Pre)Fa[To[i]][0]=Now,Dis[To[i]][0]=Len[i],Dfs(To[i],Now);
}
int Lc(int U,int V,int &Ans){
if(Dep[U]<Dep[V])swap(U,V);
Fd(i,Mx,0)if(Dep[Fa[U][i]]>=Dep[V])Ans+=Dis[U][i],U=Fa[U][i];
if(U==V)return U;
Fd(i,Mx,0)if(Fa[U][i]!=Fa[V][i])Ans+=Dis[U][i]+Dis[V][i],U=Fa[U][i],V=Fa[V][i];
Ans+=Dis[U][0]+Dis[V][0];return Fa[U][0];
}
int Up(int U,int V){
Fd(i,Mx,0)if(Two[i]<=V)V-=Two[i],U=Fa[U][i];
return U;
}
int main(){
//freopen("d.in","r",stdin);
//freopen("d.out","w",stdout);
Two[0]=1;Fo(i,1,15)Two[i]=Two[i-1]<<1;
int T;scanf("%d",&T);
Fo(cas,1,T){
scanf("%d",&N);Mx=log(N)/log(2);
Edge=0;Fo(i,1,N)Begin[i]=0;
int U,V,W;
Fo(i,2,N){
scanf("%d%d%d\n",&U,&V,&W);
Insert(U,V,W);Insert(V,U,W);
}
Dfs(1,0);int Lca;char Ch;
Fo(j,1,Mx)Fo(i,1,N)Fa[i][j]=Fa[Fa[i][j-1]][j-1],Dis[i][j]=Dis[i][j-1]+Dis[Fa[i][j-1]][j-1];
while(1){
Ch=getchar();
if(Ch=='D'){
Ch=getchar();
if(Ch=='I'){
scanf("ST%d%d\n",&U,&V);
int Ans=0;Lc(U,V,Ans);
printf("%d\n",Ans);
}else{scanf("NE\n\n");break;}

}else{
scanf("TH%d%d%d\n",&U,&V,&W);
int Ans,Lca=Lc(U,V,Ans);
if(Dep[U]-Dep[Lca]+1>=W)printf("%d\n",Up(U,W-1));
else printf("%d\n",Up(V,Dep[U]+Dep[V]-Dep[Lca]*2+1-W));
}
}
printf("\n");
}
return 0;
}