SPOJ 题目913QTREE2 - Query on a tree II（Link Cut Tree 查询路径第k个点）

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
题目大意：DIST 4 6询问4 到6的距离，KTH 4 6 4，询问4 到6经过的第4个点
ac代码

#include<stdio.h>   
#include<string.h>   
#include<stdlib.h>   
#include<queue>   
#include<iostream>   
#define MAXN 10010
using namespace std;  
int head[10010],cnt,vis[10010];  
struct s  
{  
    int u,v,w,next;  
}edge[10010<<1];  
struct LCT    
{    
    int bef[10010],pre[10010],next[10010][2],key[10010],sum[10010],belong[10010],num[10010];  
    void init()    
    {    
        memset(pre,0,sizeof(pre));  
        memset(next,0,sizeof(next));    
    }  
    void pushup(int x)  
    {  
        sum[x]=key[x]+sum[next[x][1]]+sum[next[x][0]];  
		num[x]=num[next[x][1]]+num[next[x][0]]+1;
    }  
    void rotate(int x,int kind)    
    {    
        int y,z;    
        y=pre[x];    
        z=pre[y];    
        next[y][!kind]=next[x][kind];    
        pre[next[x][kind]]=y;    
        next[z][next[z][1]==y]=x;    
        pre[x]=z;    
        next[x][kind]=y;    
        pre[y]=x;  
        pushup(y);  
    }    
    void splay(int x)    
    {    
        int rt;    
        for(rt=x;pre[rt];rt=pre[rt]);    
        if(x!=rt)    
        {    
            bef[x]=bef[rt];    
            bef[rt]=0;    
            while(pre[x])    
            {    
                if(next[pre[x]][0]==x)    
                {    
                    rotate(x,1);    
                }    
                else    
                    rotate(x,0);    
            }    
            pushup(x);  
        }    
    }   
    void access(int x)    
    {    
        int fa;    
        for(fa=0;x;x=bef[x])    
        {    
            splay(x);    
            pre[next[x][1]]=0;    
            bef[next[x][1]]=x;    
            next[x][1]=fa;    
            pre[fa]=x;    
            bef[fa]=0;    
            fa=x;    
            pushup(x);  
        }    
    }  
    int query(int x,int y)  
    {  
        access(y);  
        for(y=0;x;x=bef[x])  
        {  
            splay(x);  
            if(!bef[x])  
                return sum[y]+sum[next[x][1]];  
            pre[next[x][1]]=0;  
            bef[next[x][1]]=x;  
            next[x][1]=y;  
            pre[y]=x;  
            bef[y]=0;  
            y=x;  
            pushup(x);  
        }  
    } 
	int select(int x,int k)
	{
		while(num[next[x][0]]+1!=k)
		{
			if(num[next[x][0]]+1>k)
				x=next[x][0];
			else
			{
				k-=num[next[x][0]]+1;
				x=next[x][1];
			}
		}
		return x;
	}
	int KTH(int y,int x,int k)
	{
		access(y);  
        for(y=0;x;x=bef[x])  
        {  
            splay(x);  
            if(!bef[x])  
			{
				if(num[next[x][1]]+1==k)
					return x;
				else
					if(num[next[x][1]]+1>k)
						return select(next[x][1],num[next[x][1]]-k+1);
					return select(y,k-num[next[x][1]]-1);
			}
            pre[next[x][1]]=0;  
            bef[next[x][1]]=x;  
            next[x][1]=y;  
            pre[y]=x;  
            bef[y]=0;  
            y=x;  
            pushup(x);  
        } 
	}
}lct;
void add(int u,int v,int w)  
{  
    edge[cnt].u=u;  
    edge[cnt].v=v;  
    edge[cnt].w=w;  
    edge[cnt].next=head[u];  
    head[u]=cnt++;  
}  
void bfs(int u)  
{  
    int i,y;  
    queue<int>q;  
    memset(vis,0,sizeof(vis));  
    vis[u]=1;  
    q.push(u);  
    while(!q.empty())  
    {  
        u=q.front();  
        q.pop();  
        for(int i=head[u];i!=-1;i=edge[i].next)  
        {  
            int v=edge[i].v;  
            if(!vis[v])  
            {  
                lct.bef[v]=u;  
                lct.key[v]=lct.sum[v]=edge[i].w;  
                vis[v]=1;  
                q.push(v);  
            }  
        }  
    }  
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,i;
		scanf("%d",&n);
		cnt=0;
		memset(head,-1,sizeof(head));
		for(i=1;i<n;i++)
		{
			int u,v,w;
			scanf("%d%d%d",&u,&v,&w);
			add(u,v,w);
			add(v,u,w);
		}
		lct.init();
		bfs(1);
		char op[10];
		while(scanf("%s",op)!=EOF)
		{
			if(!strcmp(op,"DONE"))
				break;
			if(op[0]=='D')
			{
				int x,y;
				scanf("%d%d",&x,&y);
				printf("%d\n",lct.query(x,y));
			}
			else
			{
				int x,y,z;
				scanf("%d%d%d",&x,&y,&z);
				printf("%d\n",lct.KTH(x,y,z));
			}
		}
	}
}

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