hdu 2586 How far away ？【LCA】

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#hdu 2586 #杭电 2586

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本文探讨了一种解决复杂路径查询问题的高效算法，通过构建有向图并利用深度优先搜索（DFS）和并查集（Union-Find）来解决从任意房屋A到任意房屋B的距离查询。该算法在实际应用中表现出了良好的性能，适用于大规模网络的路径查询需求。

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How far away ？

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11322 Accepted Submission(s): 4121

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<k<=40000).The houses are labeled from 1 to n.
Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.

Output

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

Sample Input

3 2

1 2 10

3 1 15

1 2

2 3

2 2

1 2 100

1 2

2 1

Sample Output

100

Source

ECJTU 2009 Spring Contest

思路来源：http://blog.youkuaiyun.com/mengxiang000000/article/details/51437842

AC代码：

#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
int head[100000];
int f[100000];
int qhead[100000];
int vis[100000];
int dir[100000];
int problem[100000][2];
struct EdgeNode
{
    int from;
    int to;
    int w;
    int lca;
    int next;
} e[100000],q[100000];
int n,m,ask,cont;
int find(int a)
{
    int r=a;
    while(f[r]!=r)
        r=f[r];
    int i=a;
    int j;
    while(i!=r)
    {
        j=f[i];
        f[i]=r;
        i=j;
    }
    return r;
}
void merge(int a,int b)
{
    int A,B;
    A=find(a);
    B=find(b);
    if(A!=B)
        f[B]=A;
}
void add(int from,int to,int w)
{
    e[cont].from=from;
    e[cont].to=to;
    e[cont].w=w;
    e[cont].next=head[from];
    head[from]=cont;
    cont++;
    e[cont].from=to;
    e[cont].to=from;
    e[cont].w=w;
    e[cont].next=head[to];
    head[to]=cont;
    cont++;
}
void addq(int from,int to)
{
    q[cont].from=from;
    q[cont].to=to;
    q[cont].lca=-1;
    q[cont].next=qhead[from];
    qhead[from]=cont;
    cont++;
    q[cont].from=to;
    q[cont].to=from;
    q[cont].lca=-1;
    q[cont].next=qhead[to];
    qhead[to]=cont;
    cont++;
}
void LCA(int u)
{
    f[u]=u;
    vis[u]=1;
    for(int k=head[u]; k!=-1; k=e[k].next)
    {
        int to=e[k].to;
        if(vis[to]==0)
        {
            int w=e[k].w;
            dir[to]=dir[u]+w;
            LCA(to);
            merge(u,to);
        }
    }
    for(int k=qhead[u]; k!=-1; k=q[k].next)
    {
        int to=q[k].to;
        if(vis[to]==1)
        {
            q[k].lca=find(to);
            q[k^1].lca=q[k].lca;
        }
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        cont=0;
        for(int i=1; i<=n; i++)f[i]=i;
        memset(vis,0,sizeof(vis));
        memset(head,-1,sizeof(head));
        memset(qhead,-1,sizeof(qhead));
        memset(dir,0,sizeof(dir));
        for(int i=0; i<n-1; i++)
        {
            int x,y,w;
            scanf("%d%d%d",&x,&y,&w);
            add(x,y,w);
        }
        cont=0;
        int tot=0;
        for(int i=0; i<m; i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            addq(x,y);
        }
        LCA(1);
        for(int i=0; i<m; i++)
        {
            int tmp=i*2,u=q[tmp].from,v=q[tmp].to,lca=q[tmp].lca;
            printf("%d\n",dir[u]+dir[v]-2*dir[lca]);
        }
    }
}