hdu 2586 How far away ？(倍增法LCA)

How far away ？

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

Total Submission(s): 15643    Accepted Submission(s): 5942


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

2
3 2
1 2 10
3 1 15
1 2
2 3

2 2
1 2 100
1 2
2 1

 

Sample Output

10
25
100
100

 

  题意：有n个房子，n-1条路把它们链接起来，给出每条路的距离，有k个问题，询问两个房间的最小距离。

分析：n个点，n-1条边，组成了一颗树，所以 这是求树上两结点距离的问题，试了一手自己的倍增法求LCA模板。

之前有一篇写过倍增法求LCA 的详解  点击打开链接

AC代码：

#include<stdio.h>
#include<string.h>
#include<vector>
#include<math.h>
using namespace std;
struct tree
{
int from,to,w;
tree(int ff,int tt,int ww)
{
from=ff;to=tt;w=ww;
}
tree(){}
};
vector<tree>G;
vector<
d50f
int>V[400200];
int mxdeep;
int father[50050][30]; //记录祖先结点编号
int deep[50050]; //记录结点深度
int dis[50050][30]; //记录结点和祖先结点的距离
void addtree(int from,int to,int w) //建边
{
G.push_back(tree(from,to,w));
G.push_back(tree(to,from,w));
int e=G.size();
V[from].push_back(e-2);
V[to].push_back(e-1);
}
void init(int n)
{
memset(father,0,sizeof(father));
memset(deep,0,sizeof(deep));
memset(dis,0,sizeof(dis));
mxdeep=log(n*1.0)/log(2.0);
G.clear();
for(int i=0;i<=400000;i++)
V[i].clear();
}
void dfs(int root)
{
for(int i=1;i<=mxdeep;i++)
{
father[root][i]=father[father[root][i-1]][i-1];
dis[root][i]=dis[father[root][i-1]][i-1]+dis[root][i-1];
if(!father[root][i]) //优化，当为0的时候说明已经更新到了根结点位置
break;
}
for(int i=0;i<V[root].size();i++)
{
tree &e=G[V[root][i]];
if(e.to!=father[root][0]) //说明e.to是root的儿子
{
deep[e.to]=deep[root]+1;
father[e.to][0]=root; //建立父子关系
dis[e.to][0]=e.w; //更新到父亲的距离
dfs(e.to);
}
}
}
int lca(int u,int v)
{
int ans=0;
if(deep[u]>deep[v]) //让u结点不在v结点下方，方便操作
swap(u,v);
for(int i=mxdeep;i>=0;i--) //将 v结点上移，但不能高于u结点
{
if(deep[u]<deep[v]&&deep[u]<=deep[father[v][i]])
{
// printf("! %d %d %d %d\n",u,v,i,father[v][i]);
ans+=dis[v][i];
v=father[v][i];
}
}
if(v==u)
return ans;
for(int i=mxdeep;i>=0;i--)
{
if(father[v][i]!=father[u][i]) //一起移动到共同父亲处
{
ans+=dis[v][i]+dis[u][i];
v=father[v][i];
u=father[u][i];
}
}
if(v!=u) ans+=dis[v][0]+dis[u][0]; // 如果没有移动到同一处，可能是 1->u 1->v ，那么向上移动一个位置就可以了
return ans;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int n,k;
scanf("%d%d",&n,&k);
init(n);
int root;
for(int i=0;i<n-1;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
father[v][0]=u;
dis[v][0]=w;
addtree(u,v,w);
if(father[u][0]==0)
root=u;
}
deep[root]=1;
dfs(root); //建树
for(int i=0;i<k;i++)
{
int u,v;
scanf("%d%d",&u,&v);
printf("%d\n",lca(u,v));
}
}
}