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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, bof 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

树链剖分+线段树

维护两点建路径上的边权最大值。。。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>

using namespace std;

typedef long long int LL;

const int maxn=110000;

struct Edge
{
	int to,next;
}edge[maxn*4];

int Adj[maxn],Size;

void init_edge()
{
	memset(Adj,-1,sizeof(Adj)); Size=0;
}

void add_edge(int u,int v)
{
	edge[Size].to=v; edge[Size].next=Adj[u]; Adj[u]=Size++;
}

int fa[maxn],deep[maxn],num[maxn],son[maxn];
int top[maxn],p[maxn],rp[maxn],pos;

void init()
{
	init_edge();
	memset(son,-1,sizeof(son));
	pos=1;
}

void dfs1(int u,int pre,int d)
{
	num[u]=1; fa[u]=pre; deep[u]=d;
	for(int i=Adj[u];~i;i=edge[i].next)
	{
		int v=edge[i].to;
		if(v==pre) continue;
		dfs1(v,u,d+1);
		num[u]+=num[v];
		if(son[u]==-1||num[son[u]]<num[v])
			son[u]=v;
	}
}

void getPOS(int u,int to)
{
	top[u]=to;
	p[u]=pos++;
	rp[p[u]]=u;
	if(son[u]!=-1) getPOS(son[u],to);
	for(int i=Adj[u];~i;i=edge[i].next)
	{
		int v=edge[i].to;
		if(v!=fa[u]&&v!=son[u])
			getPOS(v,v);
	}
}

int n;
int e[maxn][3];

///segTree
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
int maxnum[maxn<<2];

void build(int l,int r,int rt)
{
    maxnum[rt]=0;
    if(l==r) return ;
    int m=(l+r)/2;
    build(lson);
    build(rson);
}

void push_up(int rt)
{
    maxnum[rt]=max(maxnum[rt<<1],maxnum[rt<<1|1]);
}

void update(int pos,int val,int l,int r,int rt)
{
    if(l==pos&&r==pos)
    {
        maxnum[rt]=val;
        return ;
    }
    int m=(l+r)/2;
    if(pos<=m) update(pos,val,lson);
    else update(pos,val,rson);
    push_up(rt);
}

int query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&r<=R)
    {
        return maxnum[rt];
    }
    int m=(l+r)/2;
    int ret=0;
    if(L<=m) ret=max(ret,query(L,R,lson));
    if(R>m) ret=max(ret,query(L,R,rson));
    return ret;
}

int find(int u,int v)
{
    int f1=top[u],f2=top[v];
    int ret=0;
    while(f1!=f2)
    {
        if(deep[f1]<deep[f2])
        {
            swap(f1,f2); swap(u,v);
        }
        ret=max(ret,query(p[f1]-1,p[u]-1,1,n,1));
        u=fa[f1]; f1=top[u];
    }
    if(u==v) return ret;
    if(deep[u]>deep[v]) swap(u,v);
    ret=max(ret,query(p[son[u]]-1,p[v]-1,1,n,1));
    return ret;
}

void showit(int l,int r,int rt)
{
    cout<<rt<<": "<<l<<" <---> "<<r<<"  max: "<<maxnum[rt]<<endl;
    if(l==r) return ;
    int m=(l+r)/2;
    showit(lson); showit(rson);
}

int main()
{
    int T_T;
    scanf("%d",&T_T);
    while(T_T--)
    {
        scanf("%d",&n);
        init();
        n--;
        for(int i=1;i<=n;i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            e[i][0]=a;e[i][1]=b;e[i][2]=c;
            add_edge(a,b);
            add_edge(b,a);
        }
        dfs1(1,1,0);
        getPOS(1,1);
        build(1,n,1);
        for(int i=1;i<=n;i++)
        {
            if(deep[e[i][0]]>deep[e[i][1]])
                swap(e[i][0],e[i][1]);
            update(p[e[i][1]]-1,e[i][2],1,n,1);
        }
        char op[10];
        while(scanf("%s",op)!=EOF)
        {
            if(op[0]=='Q')
            {
                int a,b;
                scanf("%d%d",&a,&b);
                printf("%d\n",find(a,b));
            }
            else if(op[0]=='C')
            {
                int a,b;
                scanf("%d%d",&a,&b);
                update(p[e[a][1]]-1,b,1,n,1);
            }
            else if(op[0]=='D')
            {
                putchar(10);
                break;
            }
        }
    }
	return 0;
}

SPOJ QTREE 375. Query on a tree

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原文地址：http://blog.csdn.net/ck_boss/article/details/39735563