poj 3237 树链剖分（区间更新，区间查询）

http://poj.org/problem?id=3237

Description

You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edges are numbered 1 through N − 1. Each edge is associated with a weight. Then you are to execute a series of instructions on the tree. The instructions can be one of the following forms:

CHANGE i v	Change the weight of the ith edge to v
NEGATE a b	Negate the weight of every edge on the path from a to b
QUERY a b	Find the maximum weight of edges on the path from a to b

Input

The input contains multiple test cases. The first line of input contains an integer t (t ≤ 20), the number of test cases. Then follow the test cases.

Each test case is preceded by an empty line. The first nonempty line of its contains N (N ≤ 10,000). The next N − 1 lines each contains three integers a, b and c, describing an edge connecting nodes a and bwith weight c. The edges are numbered in the order they appear in the input. Below them are the instructions, each sticking to the specification above. A lines with the word “DONE” ends the test case.

Output

For each “QUERY” instruction, output the result on a separate line.

Sample Input

1

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

Sample Output

1
3

/**
poj 3237 树链剖分（区间更新，区间查询）
题目大意：给定一棵树，动态修改：1.对于给定的两点之间的所有边权取相反数，2对于给定边修改值，动态查询：指定两点间边权最大值
解题思路：树链剖分。在线段树区间维护的时候要维护一个最大值和一个最小值，因为一翻转二者就会相互装换
*/
#include 
#include 
#include 
#include 
using namespace std;
typedef long long LL;
const int maxn=10005;
int fa[maxn],dep[maxn],son[maxn],d[maxn][3],num[maxn],top[maxn],siz[maxn];
int n,z,maxx[maxn*4],minn[maxn*4],col[maxn*4];
int head[maxn],ip;

void init()
{
    memset(col,0,sizeof(col));
    memset(head,-1,sizeof(head));
    ip=0;
}

struct note
{
    int v,w,next;
} edge[maxn*4];

void addedge(int u,int v,int w)
{
    edge[ip].v=v,edge[ip].w=w,edge[ip].next=head[u],head[u]=ip++;
}

void dfs(int u,int pre)
{
    son[u]=0,siz[u]=1,dep[u]=dep[pre]+1,fa[u]=pre;
    for(int i=head[u]; i!=-1; i=edge[i].next)
    {
        int v=edge[i].v;
        if(v==pre)continue;
        dfs(v,u);
        siz[u]+=siz[v];
        if(siz[son[u]]loc||r>1;
    update(root<<1,l,mid,loc,z);
    update(root<<1|1,mid+1,r,loc,z);
    push_up(root);
}

void update1(int root, int l,int r,int x,int y)
{
    if(l>y||r>1;
    update1(root<<1,l,mid,x,y);
    update1(root<<1|1,mid+1,r,x,y);
    push_up(root);
}

int query(int root ,int l,int r,int x,int y)
{
    if(l>y||r>1;
    return max(query(root<<1,l,mid,x,y),query(root<<1|1,mid+1,r,x,y));
}
int main()
{
    ///freopen("data.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        init();
        for(int i=1; i>1;
        z=0,siz[0]=0,dep[0]=0;
        dfs(root,0);
        init_que(root,root);
        for(int i=1; idep[d[i][1]])
            {
                swap(d[i][0],d[i][1]);
            }
            update(1,1,z,num[d[i][1]],d[i][2]);
        }
        while(1)
        {
            char s[10];
            scanf("%s",s);
            if(s[0]=='D')break;
            int x,y;
            scanf("%d%d",&x,&y);
            if(s[0]=='C')
            {
                update(1,1,z,num[d[x][1]],y);
            }
            else if(s[0]=='N')
            {
                int f1=top[x],f2=top[y];
                while(f1!=f2)
                {
                    if(dep[f1]dep[y])
                    {
                        swap(x,y);
                    }
                    update1(1,1,z,num[son[x]],num[y]);
                }
            }
            else
            {
                int f1=top[x],f2=top[y];
                int sum=-0x3f3f3f3f;
                while(f1!=f2)
                {
                    if(dep[f1]dep[y])
                    {
                        swap(x,y);
                    }
                    sum=max(query(1,1,z,num[son[x]],num[y]),sum);
                }
                printf("%d\n",sum);
            }
        }
    }
    return 0;
}