hdu2586 树上两点之间的距离 tarjan

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6650    Accepted Submission(s): 2475

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

23 21 2 103 1 151 22 32 21 2 1001 22 1

 

Sample Output

1025100100

 

此题是树上两点之间的距离，直观想法是暴力枚举树根，dfs出所有距离，n^2的复杂度无法容忍。

那么我们是否可以固定一个树根，然后计算每个点到树根的距离dist，假设a，b的LCA是c，那么a到b这条路径我们可以拆成a->c,c->b,这个信息我们显然无法得到，那么我们加上两端c->root,root->c，合并后就是a->root,root->b,也就是dist[a]+dist[b],因为我们多加了两段2*dist[c]，因此最后就是dist[a]+dist[b]-2*dist[c]。

这里可用在线RMQ算法来处理欧拉序列或者使用离线tarjan算法。

在线RMQ算法较好理解，而tarjan算法有点难理解。

其实tarjan思想的本质就是递归处理，假设我们当前在处理u的点，我们须先处理u的所有子树(这里是递归)，然后访问完u的子树，表示u已访问完毕，我们将vis[u]置1，然后立刻处理与u向关联的询问，假设询问（u，v），那么我们考察一下v是否被访问，假设vis[v]=0，那么我们不处理，因为下次当我们访问到v时再来处理（v,u）{此时u已访问}，为了保证算法正确性，我们对每个询问标记两次，（u，v）和（v,u），即保证处理到一次。而对于vis[v]=1的情况，我们要处理(u,v)的LCA，那么LCA是什么呢？画张图就知道了，一定是访问完v往上走之后下到u，因此我们需要并查集记录一下，对u每访问完一个子树，将子树与u合并，然后让该集合指向u，也就是上走到u，可以画张图，应该就能理解。

代码：

#include<iostream>#include<cstdio>#include<cstring>#define Maxn 40010#define Maxm 210using namespace std;struct edge{    int fr,to,w,lca,next;}p[Maxn<<1],ask[Maxm<<1];int head[Maxn],ah[Maxn];int tot,tot1;void addedge(int a,int b,int c){    p[tot].to=b;    p[tot].w=c;    p[tot].next=head[a];    head[a]=tot++;}void addedge1(int a,int b){    ask[tot1].fr=a;    ask[tot1].to=b;    ask[tot1].next=ah[a];    ah[a]=tot1++;}int fa[Maxn];int findset(int x){    return fa[x]==x?x:(fa[x]=findset(fa[x]));}void unionset(int a,int b){    fa[findset(a)]=findset(b);}int vis[Maxn];int anc[Maxn];int dist[Maxn];void LCA(int u,int fa){    for(int i=head[u];i!=-1;i=p[i].next){        int v=p[i].to;        if(fa!=v){            dist[v]=dist[u]+p[i].w;            LCA(v,u);            unionset(u,v); //将子树合并到父亲            anc[findset(u)]=u; //维护新集合指向父亲        }    }    vis[u]=1; //设置已访问    for(int i=ah[u];i!=-1;i=ask[i].next){ //处理与u关联的边        int v=ask[i].to;        if(vis[v]) //若v已访问,则说明u,v的lca是v所在集合的指向            ask[i].lca=ask[i^1].lca=anc[findset(v)];    }}void init(int n){    tot=tot1=0;    memset(head,-1,sizeof head);    memset(ah,-1,sizeof ah);    memset(vis,0,sizeof vis);    for(int i=1;i<=n;i++) fa[i]=i;}int main(){    int t,n,m,a,b,c;    cin>>t;    while(t--){        cin>>n>>m;        init(n);        for(int i=1;i<n;i++){            scanf("%d%d%d",&a,&b,&c);            addedge(a,b,c);            addedge(b,a,c);        }        for(int i=1;i<=m;i++){            scanf("%d%d",&a,&b);            addedge1(a,b);            addedge1(b,a);        }        dist[1]=0;        LCA(1,-1);        for(int i=0;i<tot1;i+=2)            printf("%d\n",dist[ask[i].fr]+dist[ask[i].to]-2*dist[ask[i].lca]);    }    return 0;}