【SPOJ 375】 【POJ 3237】 375. Query on a tree 树链剖分

SPOJ Problem Set (classical) 375. Query on a tree Problem code: QTREE

You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1.

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

• CHANGE i ti : change the cost of the i-th edge to ti
  or
• QUERY a b : ask for the maximum edge cost on the path from node a to node b

Input

The first line of input contains an integer t, the number of test cases (t <= 20). 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 <= 1000000),
• The next lines contain instructions "CHANGE i ti" or "QUERY a b",
• The end of each test case is signified by the string "DONE".

There is one blank line between successive tests.

Output

For each "QUERY" operation, write one integer representing its result.

Example

Input:
1

3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE

Output:
1
3

---

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>

const int INF = 233333333;

using namespace std;

struct Tree
{
	int LC, RC, FA, MAX, KEY;

};

#define lc(x) tree[x].LC
#define rc(x) tree[x].RC
#define fa(x) tree[x].FA
#define Max(x) tree[x].MAX
#define key(x) tree[x].KEY

struct Edge
{
	int next, node, v;
};

Tree tree[10010];
Edge e[20010];
int n, m, T;
int h[10010], tot;
int belong[10010];
queue<int> q;

bool vis[10010];
inline void Update(int x)
{
    Max(x) = max(max(Max(lc(x)),  Max(rc(x))),  key(x));
}

inline void set(int x)
{
    lc(x) = rc(x) = 0;
}

inline bool isRINFt(int x)
{
    return lc(fa(x)) != x && rc(fa(x)) != x;
}

inline void Zig(int x)
{
	int y=fa(x), z=fa(y);
	if (lc(z)==y) lc(z)=x;
	else if (rc(z)==y) rc(z)=x;
	fa(x)=z;
	fa(rc(x))=y;
	lc(y)=rc(x);
	fa(y)=x;
	rc(x)=y;
	Update(y);
}

inline void Zag(int x)
{
	int y=fa(x), z=fa(y);
	if (lc(z)==y) lc(z)=x;
	else if (rc(z)==y) rc(z)=x;
	fa(x)=z;
	fa(lc(x))=y;
	rc(y)=lc(x);
	fa(y)=x;
	lc(x)=y;
	Update(y);
}

inline void Splay(int x)
{
	for(int y, z;!isRINFt(x);)
	{
		y=fa(x);z=fa(y);
		if (isRINFt(y))
        {
            if (lc(y)==x)Zig(x);
			else Zag(x);
        }
		else
        {
			if (lc(z)==y)
            {
                if (lc(y)==x)Zig(y), Zig(x);
				else Zag(x), Zig(x);
            }
			else
            {
                if (rc(y)==x)Zag(y), Zag(x);
				else Zig(x), Zag(x);
            }
        }
	}
	Update(x);
}

inline void Expose(int x)
{
    for(int y=0;x;x=fa(x))
    {
        Splay(x);
        rc(x)=y;
        Update(x);
        y=x;
    }
}

void Build(void)
{
	q.push(1);vis[1]=true;
	int x, y;set(1);fa(1)=0;
	while (!q.empty())
	{
		x=q.front();q.pop();
		for (int i = h[x]; i; i = e[i].next)
		{
			y=e[i].node;
			if (vis[y])continue;
			set(y);fa(y)=x;key(y)=e[i].v;belong[i>>1]=y;vis[y]=true;q.push(y);
		}
	}
}

inline void Modify(int x, int y)
{
    key(x)=y;
    Splay(x);
}

int Query(int x, int y)
{
	Expose(y);
	for(y=0;x;x=fa(x))
	{
		Splay(x);
		if (!fa(x)){return max(Max(rc(x)), Max(y));}
		rc(x)=y;Update(x);y=x;
	}
}

inline void add(int a, int b, int c)
{
    e[++tot].next=h[a];
    e[tot].node=b;
    e[tot].v=c;
    h[a]=tot;
}

inline void Work(void)
{
	char ch[10];int a, b;
	while (1)
	{
		scanf("%s", ch);
		if (ch[0]=='D') return;
		scanf("%d%d", &a, &b);
		if (ch[0]=='C') Modify(belong[a], b);
		else printf("%d\n", Query(a, b));
	}
}
int main(void)
{
	for(scanf("%d", &T); T; T--)
	{
		scanf("%d", &n);
		int a, b, c;
		tot = 1;
        for (int i = 0; i <= n; ++i) h[i]=vis[i]=0;
        for (int i = 1; i <= n - 1; ++i) scanf("%d%d%d", &a, &b, &c), add(a, b, c), add(b, a, c);
        key(0)=Max(0)=-INF;
        Build();
		Work();
	}
	return 0;
}