SPOJ QTREE - Query on a tree（树链剖分）

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:131 2 12 3 2QUERY 1 2CHANGE 1 3QUERY 1 2DONEOutput:1
3
分析：入门题。细节还要看下。
#include<cstdio>#include<cstring>#include<algorithm>#include<vector>#include<string>#include<iostream>#include<queue>#include<cmath>#include<map>#include<stack>#include<set>using namespace std;#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )#define CLEAR( a , x ) memset ( a , x , sizeof a )const int INF=0x3f3f3f3f;typedef long long LL;const int maxn=10010;struct node{    int to,next;}e[maxn*2];int head[maxn],tot,pos;int top[maxn],fa[maxn],deep[maxn],num[maxn],p[maxn],son[maxn],fp[maxn];int t,n;int op[maxn][3];void init(){    tot=0;pos=0;    CLEAR(head,-1);    CLEAR(son,-1);}void addedge(int u,int v){    e[tot].to=v;e[tot].next=head[u];    head[u]=tot++;}void dfs1(int u,int pre,int d){    deep[u]=d;fa[u]=pre;    num[u]=1;    for(int i=head[u];i!=-1;i=e[i].next)    {        int v=e[i].to;        if(pre==v) continue;        dfs1(v,u,d+1);        num[u]+=num[v];        if(son[u]==-1||num[v]>num[son[u]])            son[u]=v;    }}void dfs2(int u,int d){    top[u]=d;    p[u]=pos++;    fp[p[u]]=u;    if(son[u]==-1) return ;    dfs2(son[u],d);    for(int i=head[u];i!=-1;i=e[i].next)    {        int v=e[i].to;        if(son[u]!=v&&v!=fa[u])            dfs2(v,v);    }}int sum[maxn<<2];void pushup(int rs){    sum[rs]=max(sum[rs<<1],sum[rs<<1|1]);}void update(int x,int c,int l,int r,int rs){    if(l==r)    {        sum[rs]=c;        return ;    }    int mid=(l+r)>>1;    if(x<=mid) update(x,c,l,mid,rs<<1);    else  update(x,c,mid+1,r,rs<<1|1);    pushup(rs);}int query(int x,int y,int l,int r,int rs){    if(l>=x&&r<=y)        return sum[rs];    int mid=(l+r)>>1;    int res=0;    if(x<=mid) res=max(res,query(x,y,l,mid,rs<<1));    if(y>mid)  res=max(res,query(x,y,mid+1,r,rs<<1|1));    return res;}int Find(int u,int v){    int f1=top[u],f2=top[v];    int res=0;    while(f1!=f2)    {        if(deep[f1]<deep[f2])        {            swap(f1,f2);            swap(u,v);        }        res=max(res,query(p[f1],p[u],1,pos,1));        u=fa[f1];f1=top[u];    }    if(u==v) return res;    if(deep[u]>deep[v]) swap(u,v);    return max(res,query(p[son[u]],p[v],1,pos,1));}int main(){    int x,y,w;    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        init();        for(int i=0;i<n-1;i++)        {            scanf("%d%d%d",&op[i][0],&op[i][1],&op[i][2]);            addedge(op[i][0],op[i][1]);            addedge(op[i][1],op[i][0]);        }        dfs1(1,0,0);        dfs2(1,1);        CLEAR(sum,0);//build(1,1,n);        for(int i=0;i<n-1;i++)        {            if(deep[op[i][0]]>deep[op[i][1]])                swap(op[i][0],op[i][1]);            update(p[op[i][1]],op[i][2],1,pos,1);        }        char str[10];        while(~scanf("%s",str))        {            if(str[0]=='D')                break;            scanf("%d%d",&x,&y);            if(str[0]=='Q')               printf("%d\n",Find(x,y));            else               update(p[op[x-1][1]],y,1,pos,1);        }    }    return 0;}