spoj375 Query on a tree(树链剖分入门题)

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <cstdio>
using namespace std;
const int MAXN=10010;
struct Edge
{
    int to,next;
}edge[MAXN*2];
int head[MAXN],tot;
int top[MAXN];//top[v]表示v所在的重链的顶端节点
int fa[MAXN];//父亲节点
int deep[MAXN];//深度
int num[MAXN];//num[v]表示以v为根的子树的节点数
int p[MAXN];//p[v]表示v与其父亲节点的连边在线段树中的位置
int fp[MAXN];//和p数组相反
int son[MAXN];//重儿子
int pos;
void init()
{
    tot=0;
    memset(head,-1,sizeof(head));
    pos=0;
    memset(son,-1,sizeof(son));
}
void addedge(int u,int v)
{
    edge[tot].to=v;
    edge[tot].next=head[u];
    head[u]=tot++;
}
void dfs1(int u,int pre,int d)//第一遍dfs求出fa,deep,num,son
{
    deep[u]=d;
    fa[u]=pre;
    num[u]=1;
    for(int i=head[u]; i!=-1; i=edge[i].next)
    {
        int v=edge[i].to;
        if(v!=pre)
        {
            dfs1(v,u,d+1);
            num[u]+=num[v];
            if(son[u]==-1||num[v]>num[son[u]]) son[u]=v;
        }
    }
}
void getpos(int u,int sp)//第二遍dfs求出top和p
{
    top[u]=sp;
    if(son[u]!=-1)
    {
        p[u]=pos++;
        fp[p[u]]=u;
        getpos(son[u],sp);
    }
    else
    {
        p[u]=pos++;
        fp[p[u]]=u;
        return;
    }
    for(int i=head[u]; i!=-1; i=edge[i].next)
    {
        int v=edge[i].to;
        if(v!=son[u]&&v!=fa[u]) getpos(v,v);
    }
}
//线段树
struct Node
{
    int l,r,Max;
}segTree[MAXN*3];
void build(int i,int l,int r)
{
    segTree[i].l=l;
    segTree[i].r=r;
    segTree[i].Max=0;
    if(l==r) return;
    int mid=(l+r)/2;
    build(i<<1,l,mid);
    build((i<<1)|1,mid+1,r);
}
void push_up(int i)
{
    segTree[i].Max=max(segTree[i<<1].Max,segTree[(i<<1)|1].Max);
}
void update(int i,int k,int val)//更新线段树的第k个值为val
{
    if(segTree[i].l==k&&segTree[i].r==k)
    {
        segTree[i].Max=val;
        return;
    }
    int mid=(segTree[i].l+segTree[i].r)/2;
    if(k<=mid) update(i<<1,k,val);
    else update((i<<1)|1,k,val);
    push_up(i);
}
int query(int i,int l,int r)//查询线段树中[l,r]的最大值
{
    if(segTree[i].l==l&&segTree[i].r==r) return segTree[i].Max;
    int mid=(segTree[i].l+segTree[i].r)/2;
    if(r<=mid) return query(i<<1,l,r);
    else if(l>mid) return query((i<<1)|1,l,r);
    else return max(query(i<<1,l,mid),query((i<<1)|1,mid+1,r));
}
int find(int u,int v)//查询u->v边的最大值
{
    int f1=top[u],f2=top[v],tmp=0;
    while(f1!=f2)
    {
        if(deep[f1]<deep[f2])
        {
            swap(f1,f2);
            swap(u,v);
        }
        tmp=max(tmp,query(1,p[f1],p[u]));
        u=fa[f1];
        f1=top[u];
    }
    if(u==v) return tmp;
    if(deep[u]>deep[v]) swap(u,v);
    return max(tmp,query(1,p[son[u]],p[v]));
}
int e[MAXN][3];
int main()
{
    //freopen("in.txt","r",stdin);
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        init();
        scanf("%d",&n);
        for(int i=0; i<n-1; i++)
        {
            scanf("%d%d%d",&e[i][0],&e[i][1],&e[i][2]);
            addedge(e[i][0],e[i][1]);
            addedge(e[i][1],e[i][0]);
        }
        dfs1(1,0,0);
        getpos(1,1);
        build(1,0,pos-1);
        for(int i=0;i<n-1;i++)
        {
            if(deep[e[i][0]]>deep[e[i][1]]) swap(e[i][0],e[i][1]);
            update(1,p[e[i][1]],e[i][2]);
        }
        char op[10];
        int u,v;
        while(scanf("%s",op))
        {
            if(op[0]==‘D‘) break;
            scanf("%d%d",&u,&v);
            if(op[0]==‘Q‘)
                printf("%d\n",find(u,v));//查询u->v路径上边权的最大值
            else update(1,p[e[u-1][1]],v);//修改第u条边的长度为v
        }
    }
    return 0;
}
时间: 2024-08-10 23:48:48

spoj375 Query on a tree(树链剖分入门题)的相关文章

SPOJ375 Query on a tree 树链剖分

SPOJ375  Query on a tree   树链剖分 no tags 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

SPOJ Query on a tree 树链剖分 水题

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 tior QUERY a b : ask fo

spoj Query on a tree(树链剖分模板题)

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

SPOJ - QTREE 375 Query on a tree 树链剖分+线段树

操作1:修改第k条边权. 操作2:询问两点间最大边权. 树链剖分,然后线段树维护最大值 #include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> #include<set> #include<map> #include<queue> #include<vector> #inclu

SPOJ QTREE Query on a tree ——树链剖分 线段树

[题目分析] 垃圾vjudge又挂了. 树链剖分裸题. 垃圾spoj,交了好几次,基本没改动却过了. [代码](自带常数,是别人的2倍左右) #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define maxn 20005 int T,n,fr[maxn],h[maxn],to[maxn],ne[maxn]

spoj 375 Query on a tree (树链剖分)

Query on a tree 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 Q

spoj 375 QTREE - Query on a tree 树链剖分

题目链接 给一棵树, 每条边有权值, 两种操作, 一种是将一条边的权值改变, 一种是询问u到v路径上最大的边的权值. 树链剖分模板. #include <iostream> #include <vector> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <map> #include <set&g

SPOJ QTREE Query on a tree --树链剖分

题意:给一棵树,每次更新某条边或者查询u->v路径上的边权最大值. 解法:做过上一题,这题就没太大问题了,以终点的标号作为边的标号,因为dfs只能给点分配位置,而一棵树每条树边的终点只有一个. 询问的时候,在从u找到v的过程中顺便查询到此为止的最大值即可. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <cmath&

Query on a tree 树链剖分 [模板]

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 f