hdu 2586 How far away？【LCA】解题报告

最新推荐文章于 2024-08-26 15:30:00 发布

Hawo11

最新推荐文章于 2024-08-26 15:30:00 发布

阅读量404

点赞数 2

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/Hawo11/article/details/75007661

————单个题目——— 同时被 3 个专栏收录

222 篇文章

订阅专栏

————数据结构————

40 篇文章

订阅专栏

LCA

9 篇文章

订阅专栏

本文介绍了一种解决最近公共祖先问题的有效方法——LCA算法，并通过一个具体的村庄路径查询问题进行阐述。文章首先构建了一个由n个节点组成的树形结构，然后针对m次查询，计算任意两点间的最短距离。通过预先计算每个节点到树根的距离，结合LCA算法快速得出两点间距离。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Problem Description

There are n houses in the village and some bidirectional roads connecting them. Every day peole always like to ask like this “How far is it if I want to go from house A to house B”? Usually it hard to answer. But luckily int this village the answer is always unique, since the roads are built in the way that there is a unique simple path(“simple” means you can’t visit a place twice) between every two houses. Yout task is to answer all these curious people.

Input

First line is a single integer T(T<=10), indicating the number of test cases.
For each test case,in the first line there are two numbers n(2<=n<=40000) and m (1<=m<=200),the number of houses and the number of queries. The following n-1 lines each consisting three numbers i,j,k, separated bu a single space, meaning that there is a road connecting house i and house j,with length k(0

Output

For each test case,output m lines. Each line represents the answer of the query. Output a bland line after each test case.

思路

LCA

题目大意：一个村子里有n个房子，这n个房子用n-1条路连接起来，接下了有m次询问，每次询问两个房子a,b之间的距离是多少。

很明显的最近公共祖先问题，先建一棵树，然后求出每一点i到树根的距离dis[i],然后每次询问a，b之间的距离=dis[a]+dis[b]-2*dis[LCA(a,b)];

LCA(a,b)即是a，b的最近公共祖先。。

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<vector>
using namespace std;
const int N=40000+5;
const int M=200+5;
struct node
{
    int u,v,next,w;
}edge[2*N];
struct node1
{
    int u,v,next,num;
}edge1[2*M];
int T,num,num1,n,m;
int head[N],head1[N],father[N],dis[N],LCA[M],visit[N];
void build(int u,int v,int w)
{
    edge[num].u=u;
    edge[num].v=v;
    edge[num].w=w;
    edge[num].next=head[u];
    head[u]=num++;
}
void build1(int u,int v,int w)
{
    edge1[num1].u=u;
    edge1[num1].v=v;
    edge1[num1].num=w;
    edge1[num1].next=head1[u];
    head1[u]=num1++;
}
int getfather(int x)
{
    if (x!=father[x]) father[x]=getfather(father[x]);
    return father[x];
}
void tarjan(int u)
{
    int v;
    visit[u]=1;
    father[u]=u;
    for (int j=head1[u];j!=-1;j=edge1[j].next)
    {
        v=edge1[j].v;
        if (visit[v]) LCA[edge1[j].num]=getfather(v);
    }
    for (int j=head[u];j!=-1;j=edge[j].next)
    {
        v=edge[j].v;
        if (!visit[v]) 
        {
            dis[v]=dis[u]+edge[j].w;
            tarjan(v);
            father[v]=u;
        }
    }
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        num=0,num1=0;
        memset(head,-1,sizeof(head));
        for (int i=1;i<n;i++)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            build(a,b,c);
            build(b,a,c);
        }
        memset(visit,0,sizeof(visit));
        memset(head1,-1,sizeof(head1));
        for (int i=1;i<=m;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            build1(a,b,i);
            build1(b,a,i);
        }
        dis[1]=0;
        tarjan(1);
        for (int i=0;i<num1;i+=2)
        {
            int a,b,c;
            a=edge1[i].u;
            b=edge1[i].v;
            c=edge1[i].num;
            printf("%d\n",dis[a]+dis[b]-2*dis[LCA[c]]);
        }
    }
    return 0;
}