SPOJ-913: Query on a tree II（倍增LCA）

LCA

Description

You are given a tree (an undirected acyclic connected graph) with N nodes, and edges numbered 1, 2, 3...N-1. Each edge has an integer value assigned to it, representing its length.

We will ask you to perfrom some instructions of the following form:

• DIST a b: ask for the distance between node a and node b
  or
• KTH a b k : ask for the k-th node on the path from node a to node b
  Example:
  N = 6
  1 2 1 // edge connects node 1 and node 2 has cost 1
  2 4 1
  2 5 2
  1 3 1
  3 6 2

Path from node 4 to node 6 is 4 -> 2 -> 1 -> 3 -> 6
DIST 4 6 : answer is 5 (1 + 1 + 1 + 2 = 5)
KTH 4 6 4 : answer is 3 (the 4-th node on the path from node 4 to node 6 is 3)

Input

The first line of input contains an integer t, the number of test cases (t <= 25). t test cases follow.

For each test case:

• In the first line there is an integer N (N <= 10000)
• In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c <= 100000)
• The next lines contain instructions "DIST a b" or "KTH a b k"
• The end of each test case is signified by the string "DONE".
  There is one blank line between successive tests.

Output

For each "DIST" or "KTH" operation, write one integer representing its result.

Print one blank line after each test.

Example

Input
1
6
1 2 1
2 4 1
2 5 2
1 3 1
3 6 2
DIST 4 6
KTH 4 6 4
DONE

Output
5
3

题意： t组数据，给出一个无向图，DIST询问两点之间的距离，KTH询问两点之间的第k个点，DONE结束。

分析： 两点在图上的距离，连线一定经过他们的最近公共祖先（LCA），因此两点之间的距离即为dis[u]+dis[v]-2*dis[LCA(u,v)]；
找u，v之间的第k个点，首先要知道该点是在LCA(u,v)的左边还是右边，然后以u点或者v点向上倍增得到第k个点。
为什么不以LCA(u,v)向下找？向下可能会遇到很多分支，而以一点向上找只有一条路径。

• 代码：

#include<stdio.h>
#include<vector>
#include<algorithm>
#include<string.h>
#define maxn 10005
using namespace std;
int n;
int dep[maxn]; //深度
int dis[maxn];
int fa[maxn][15];//2^14>10000，各点向上跳2^i到达的点
struct node
{
    int en,val;
    node(int enn=0,int vall=0)
    {
        en=enn;
        val=vall;
    }
};
vector<node> V[maxn];
void clr()
{
    memset(dep,0,sizeof(dep));
    memset(dis,0,sizeof(dis));
    memset(fa,0,sizeof(fa));
    for(int i=0;i<=n;i++)
        V[i].clear();
}
void dfs(int u,int f,int d)//深度
{
    fa[u][0]=f;
    for(int i=1;i<15;i++)
        fa[u][i]=fa[fa[u][i-1]][i-1];
    dep[u]=d;
    for(int i=0;i<V[u].size();i++)
    {
        node v=V[u][i];
        if(v.en==f) continue;
        dis[v.en]=dis[u]+v.val;
        dfs(v.en,u,d+1);
    }
}
int LCA(int u,int v)
{
    if(dep[u]>dep[v])
        swap(u,v); //u浅v深
    for(int i=0;i<15;i++) //使等深度
        if((dep[v]-dep[u])>>i&1)
        v=fa[v][i];
    if(u==v) return u;
    for(int i=14;i>=0;i--) //同时向上跳
    {
        if(fa[u][i]!=fa[v][i])
        {
            u=fa[u][i];
            v=fa[v][i];
        }
    }
    return fa[u][0];
}
int kth(int u,int v,int k)
{
    int ans;
    int lca=LCA(u,v);
    int cnt=dep[u]+dep[v]-2*dep[lca]+1; //u到v数字个数
    if(dep[u]-dep[lca]+1==k)
        return lca;
    else if(dep[u]-dep[lca]+1<k)//在lca右边
    {
        k=cnt-k;//跳的步数
        u=v;
    }
    else k--;//左边
    for(int i=0;i<15;i++)
        if(k&(1<<i))
            u=fa[u][i];
    return u;
}
int main()
{
    int t;
    int a,b,c;
    char s[9];
    scanf("%d",&t);
    while(t--)
    {
        clr();
        scanf("%d",&n);
        for(int i=1;i<n;i++)
        {
            scanf("%d%d%d",&a,&b,&c);
            V[a].push_back(node(b,c));
            V[b].push_back(node(a,c));
        }
        dfs(1,0,1);
        while(1)
        {
            scanf("%s",s);
            if(s[1]=='I')
            {
                scanf("%d%d",&a,&b);
                printf("%d\n",dis[a]+dis[b]-2*dis[LCA(a,b)]);
            }
            else if(s[1]=='T')
            {
                int ans;
                scanf("%d%d%d",&a,&b,&c);
                printf("%d\n",kth(a,b,c));       
            }
            else break;
        }
    }
    return 0;
}