2014多校5（1006）hdu4916(树形DP)

最新推荐文章于 2020-01-30 14:58:56 发布

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最新推荐文章于 2020-01-30 14:58:56 发布

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分类专栏：树形DP 文章标签： ACM hdu

本文链接：https://blog.youkuaiyun.com/cq_phqg/article/details/38403333

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本文解析了一道关于树形结构的最小值查询算法题目，通过定义特定的路径最小值来解决问题。文章详细介绍了如何利用树形DP进行有效求解，并提供了实现思路及关键代码片段。

Count on the path

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 255 Accepted Submission(s): 59

Problem Description

bobo has a tree, whose vertices are conveniently labeled by 1,2,…,n.

Let f(a,b) be the minimum of vertices not on the path between vertices a and b.

There are q queries (u_i,v_i) for the value of f(u_i,v_i). Help bobo answer them.

Input

The input consists of several tests. For each tests:

The first line contains 2 integers n,q (4≤n≤10⁶,1≤q≤10⁶). Each of the following (n - 1) lines contain 2 integers a_i,b_i denoting an edge between vertices a_i and b_i (1≤a_i,b_i≤n). Each of the following q lines contains 2 integer u′_i,v′_i (1≤u_i,v_i≤n).

The queries are encrypted in the following manner.

u₁=u′₁,v₁=v′₁.
For i≥2, u_i=u′_i⊕f(u_{i - 1},v_{i - 1}),v_i=v′_i⊕f(u_i-1,v_i-1).

Note ⊕ denotes bitwise exclusive-or.

It is guaranteed that f(a,b) is defined for all a,b.

The task contains huge inputs. `scanf` in g++ is considered too slow to get accepted. You may (1) submit the solution in c++; or (2) use hand-written input utilities.

Output

For each tests:

For each queries, a single number denotes the value.

Sample Input

Sample Output

4
3
1

题意：查询不在a和b路径上标号最小的点

思路：首先以1节点为根将整颗树定形，分路径经过节点1和不经过讨论

1.路径不经过节点1，则节点1就是答案

2.路径经过节点1，那么令path[u]为不在u到1节点路径上的最小标号节点，那么在u和v两种情况中取较小的即可

处理起来有点麻烦，看代码吧~

child[u][3]表示u的前三小的子树

mintree[u]表示以u为根的子树中节点的最小标号

vis[u]表示u在1的哪颗子树中

belong[u]表示u所属于的1的子树中最小节点的标号

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define maxn 1000010
#define INF 2000000
int n,q1;
int head[maxn];
int edge[maxn<<1];
int next[maxn<<1];
int q[maxn];
int fa[maxn];
int mintree[maxn];
int child[maxn][3];
int path[maxn];
int vis[maxn];
int belong[maxn];
int d;

int mini(int x,int y)
{
    return x=0;i--)
    {
        int o=q[i];
        mintree[o]=mini(o,child[o][0]);
        child[fa[o]][2]=mintree[o];
        sort(child[fa[o]],child[fa[o]]+3);
    }

    rear=0;
    front=0;
    int dd=0;

    vis[1]=-1;

    for(i=head[1];i!=-1;i=next[i]){
        path[edge[i]]=INF;
        q[rear++]=edge[i];
        vis[edge[i]]=dd++;
        belong[edge[i]]=mintree[edge[i]];
    }

    while(front