您的位置：首页 > 其它

How far away ？+hdu+最近公共祖先

2014-09-30 09:28 337 查看

How far away ？

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5713 Accepted Submission(s): 2153

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

Sample Output

10
25
100
100
解决方案：最近公共祖先，可记录每个点到根的距离，然后两点之间的最短距离就是dist[u]+dist[v]-2*dist[LCA(u,v)]；code：#include <iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn=50000;
int ehead[maxn],qhead[maxn],fa[maxn];
int F[maxn],T[maxn];
int du[maxn];
int dist[maxn];
bool vis[maxn];
int k,n,m;
struct edge
{
int to,next;
int dis;
} E[maxn*2],Q[maxn*2];
void addedge(int from,int to,int v)
{
E[k].to=to;
E[k].dis=v;
E[k].next=ehead[from];
ehead[from]=k++;
}
void addquery(int from,int to)
{
Q[k].to=to;
Q[k].dis=0;
Q[k].next=qhead[from];
qhead[from]=k++;

}
int Root(int u)
{
return u!=fa[u]?fa[u]=Root(fa[u]):u;
}
void Unoin(int u,int v)
{
int us=Root(u);
int vs=Root(v);
if(us!=vs)
{
fa[vs]=us;
}

}
void tarjan(int u)
{

vis[u]=true;
for(int i=qhead[u]; i!=-1; i=Q[i].next)
{
int v=Q[i].to;
if(vis[v])
{
int LCA=Root(v);
Q[i].dis=dist[u]+dist[v]-2*dist[LCA];
for(int j=qhead[v]; j!=-1; j=Q[j].next)
{
if(Q[j].to==u)
{
Q[j].dis=dist[u]+dist[v]-2*dist[LCA];
break;
}
}
}
}
for(int i=ehead[u]; i!=-1; i=E[i].next)
{
int v=E[i].to;
if(vis[v]) continue;
dist[v]=dist[u]+E[i].dis;
tarjan(v);
Unoin(u,v);
}
}
void init()
{
for(int i=1; i<=n; i++)
{
fa[i]=i;
}
memset(vis,0,sizeof(vis));
memset(ehead,-1,sizeof(ehead));
memset(qhead,-1,sizeof(qhead));
memset(dist,0,sizeof(dist));
memset(du,0,sizeof(du));
k=0;
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
init();
for(int i=0; i<n-1; i++)
{
int from,to,v;
scanf("%d%d%d",&from,&to,&v);
du[to]++;
addedge(from,to,v);
addedge(to,from,v);
}
k=0;
for(int i=0; i<m; i++)
{
scanf("%d%d",&F[i],&T[i]);
addquery(F[i],T[i]);
addquery(T[i],F[i]);
}
for(int i=1; i<=n; i++)
{
if(du[i]==0)
{    //cout<<i<<endl;
tarjan(i);
}
}
for(int i=0; i<m; i++)
{
for(int v=qhead[F[i]]; v!=-1; v=Q[v].next)
{
int to=Q[v].to;
if(to==T[i])
{
printf("%d\n",Q[v].dis);
break;
}
}
}

}
return 0;
}

内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理

标签：

相关文章推荐

新的分享

章节导航