【spoj】【QTREE2 - Query on a tree II】

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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;}

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