HDU2586(LCA应用:在带权树中求任意两点之间的距离)

How far away ？

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10412 Accepted Submission(s): 3777

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

模板题.离线算法:dfs+并查集.

/*
    2586    31MS    10288K    1637 B    G++
*/
#include"cstdio"
#include"cstring"
#include"vector"
using namespace std;
const int MAXN=40005;
typedef pair<int,int> P;
vector<P> G[MAXN];
vector<P> que[MAXN];
int par[MAXN];
int fnd(int x)
{
    if(par[x]==x)
        return x;
    par[x]=fnd(par[x]);
}
int vis[MAXN];
int d[MAXN];
int ans[MAXN];
void dfs(int u,int fa)
{
    par[u]=u;
    for(int i=0;i<que[u].size();i++)
    {
        P no=que[u][i];
        if(vis[no.first])    ans[no.second]=d[u]+d[no.first]-2*d[fnd(no.first)];
    }
    
    vis[u]=1;
    for(int i=0;i<G[u].size();i++)
    {
        P now=G[u][i];
        if(now.first==fa)    continue;
        d[now.first]=d[u]+now.second;
        dfs(now.first,u);
        par[now.first]=u;
    }
    
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        memset(vis,0,sizeof(vis));
        memset(d,0,sizeof(d));
        memset(ans,0,sizeof(ans));
        int V,Q;
        scanf("%d%d",&V,&Q);
        for(int i=0;i<=V;i++)
        {
            G[i].clear();
            que[i].clear();
        }
        for(int i=1;i<=V-1;i++)
        {
            int u,v,cost;
            scanf("%d%d%d",&u,&v,&cost);
            G[u].push_back(P(v,cost));
            G[v].push_back(P(u,cost));
        }    
        for(int i=1;i<=Q;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            que[u].push_back(P(v,i));
            que[v].push_back(P(u,i));
        }
        dfs(1,-1);
        for(int i=1;i<=Q;i++)
        {
            printf("%d
",ans[i]);
        }
    //    printf("
");
    }
    return 0;
}

因为查询较少,朴素的求LCA算法就能过。

/*
        Accepted    2586    62MS    7148K    1444B    G++
*/
#include"cstdio"
#include"cstring"
#include"vector"
using namespace std;
const int MAXN=40005;
typedef pair<int,int> P;
vector<P> G[MAXN];
int depth[MAXN];
int parent[MAXN];
void dfs(int u,int fa,int d)
{
    depth[u]=d;
    parent[u]=fa;
    for(int i=0;i<G[u].size();i++)
    {
        P now=G[u][i];
        if(now.first!=fa)    dfs(now.first,u,d+1);
    }
}
void Swap(int &a,int &b)
{
    int t=a;
    a=b;
    b=t;
}
int LCA(int u,int v)
{
    int d=0;
    if(depth[u]<depth[v])    Swap(u,v);
    while(depth[u]>depth[v])
    {
        int fa=parent[u];
        for(int i=0;i<G[fa].size();i++)
        {
            P now=G[fa][i];
            if(now.first==u)
            {
                d+=now.second;
                break;
            }
        }
        u=fa;
    }
    while(u!=v)
    {
        int fa=parent[u];
        for(int i=0;i<G[fa].size();i++)
        {
            P now=G[fa][i];
            if(now.first==u)
            {
                d+=now.second;
                break;
            }
        }
        u=fa;
        
        fa=parent[v];
        for(int i=0;i<G[fa].size();i++)
        {
            P now=G[fa][i];
            if(now.first==v)
            {
                d+=now.second;
                break;
            }
        }
        v=fa;
    }
    return d;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int V,Q;
        scanf("%d%d",&V,&Q);
        for(int i=0;i<=V;i++)    G[i].clear();
        
        for(int i=1;i<=V-1;i++)
        {
            int u,v,cost;
            scanf("%d%d%d",&u,&v,&cost);
            G[u].push_back(P(v,cost));
            G[v].push_back(P(u,cost));
        }    
        dfs(1,-1,0);
        while(Q--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            printf("%d
",LCA(u,v));
        }
    }
    return 0;
}