SPOJ QTREE - Query on a tree

最新推荐文章于 2018-10-30 20:49:00 发布

weixin_30849403

最新推荐文章于 2018-10-30 20:49:00 发布

阅读量34

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/Enceladus/p/5239959.html

本文深入探讨了树链剖分和LCT（Link Cut Tree）算法的应用，通过实例展示了如何解决树状结构中的路径查询与边修改问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

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


本来应该写树链剖分的，但既然要练LCT就敲敲咯……也是调了蛮久

#include<cstdio>
#include<cstring>
#include<algorithm>
#define MN 10010
using namespace std;

int p,ca,f;
inline int read(){
    p=0;ca=getchar();f=1;
    while(ca<'0'||ca>'9') {if (ca=='-') f=-1;ca=getchar();}
    while(ca>='0'&&ca<='9') p=p*10+ca-48,ca=getchar();
    return p*f;
}
struct na{
    int y,ne,c,nu;
}b[MN*2];
int fa[MN],n,t,x,y,c,l[MN],r[MN],num,id[MN],key[MN],ch[MN][2],ma[MN];
bool rt[MN];
inline int max(int a,int b){return a>b?a:b;}
inline void update(int x){
    ma[x]=max(max(ma[ch[x][0]],ma[ch[x][1]]),key[x]);
}
inline void rot(int x){
    int y=fa[x],kind=ch[y][1]==x;
    fa[x]=fa[y];
    fa[y]=x;
    ch[y][kind]=ch[x][!kind];
    fa[ch[y][kind]]=y;
    ch[x][!kind]=y;
    if(rt[y]) rt[y]=0,rt[x]=1;else ch[fa[x]][ch[fa[x]][1]==y]=x;
    update(y);update(x);
}
inline void splay(int x){
    while(!rt[x]){
        if (rt[fa[x]]) rot(x);else
        if ((ch[fa[fa[x]]][1]==fa[x])==(ch[fa[x]][1]==x)) rot(fa[x]),rot(x);else rot(x),rot(x);
    }
}
inline void acc(int u){
    int x=0;
    while(u){
        splay(u);
        rt[ch[u][1]]=1;rt[ch[u][1]=x]=0;
        update(u);
        u=fa[x=u];
    }
}
inline void change(int x,int c){
    acc(x);
    key[x]=c;
    update(x);
}
inline void lca(int &u,int &v){
    acc(v);v=0;
    while(u){
        splay(u);
        if (!fa[u]) break;
        rt[ch[u][1]]=1;
        rt[ch[u][1]=v]=0;
        update(u);
        u=fa[v=u];
    }
}
inline int qu(int x,int y){
    lca(x,y);
    return max(ma[y],ma[ch[x][1]]);
}
inline void in(int x,int y,int c,int nu){
    num++;
    if (!l[x]) l[x]=num;else b[r[x]].ne=num;
    b[num].y=y;b[num].c=c;b[num].ne=0;b[num].nu=nu;r[x]=num;
}
inline void dfs(int x){
    for (int i=l[x];i;i=b[i].ne)
    if (!fa[b[i].y]){
        fa[b[i].y]=x;
        id[b[i].nu]=b[i].y;
        key[b[i].y]=b[i].c;
        dfs(b[i].y);
    }
}
char ss[10];
int main(){
    t=read();
    ma[0]=-1e9;
    while(t--){
        num=0;
        memset(rt,1,sizeof(rt));
        memset(fa,0,sizeof(fa));
        memset(ch,0,sizeof(ch));
        memset(l,0,sizeof(l));
        n=read();
        for (int i=1;i<n;i++){
            x=read();y=read();c=read();
            in(x,y,c,i);
            in(y,x,c,i);
        }
        fa[1]=-1;
        dfs(1);
        fa[1]=0;
        for(;;){
            scanf("%s",ss);
            if (ss[0]=='D') break;
            x=read();y=read();
            if (ss[0]=='Q') printf("%d\n",qu(x,y));else
            change(id[x],y);
        }
    }
}