最近公共祖先问题（ST算法）
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本篇主要讲解如何采用解决RMQ问题的ST算法来解决LCA问题
步骤如下：
1 DFS遍历
DFS序用ver[maxn]保存（包含回溯时经过的点）
用first[maxn]保存每一个点第一次被访问时候的位置（即ver值）
用depth[maxn]保存每一个点的深度（包括回溯时经过的点）
check[maxn]标记一个点是否被访问过
dis[maxn]保存每一个点到根节点的距离
2 ST算法
Min[maxn][maxm]保存区间使得depth[x]最小的x的值
3 RMQ（访问）
访问得到相应区间使得depth[x]最小的x的值
4 LCA求出最近公共祖先
两个点的LCA一定是两个点在DFS序中（第一次出现）出现的位置之间深度最小的那个点
5 求出两点间的距离
dis[u]+dis[v]-2*dis[lca]
附上一道裸的LCA问题 hdu2586点击打开链接
How far away ？
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 19581 Accepted Submission(s): 7673
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 Next m lines each has distinct integers i and j, you areato answer the distance between house i and house j.
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
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
#include
#include
#include
#include
#define maxn 40005
using namespace std;
int t;
int n,m;
int a,b,c;
struct node
{
int s;
int e;
int w;
int next;
}edge[maxn*2];
int head[maxn];
bool check[maxn];
int ver[maxn*2];//储存DFS序
int tot;//访问序号
int first[maxn];//每一个点第一次被访问的时候的位置
int dis[maxn];//每一点到根结点的距离
int depth[maxn*2];//第X位置的点的深度
int Min[maxn*2][18];
void add(int s,int e,int w,int &k)
{
edge[k].s=s;
edge[k].e=e;
edge[k].w=w;
edge[k].next=head[s];
head[s]=k++;
swap(s,e);
edge[k].s=s;
edge[k].e=e;
edge[k].w=w;
edge[k].next=head[s];
head[s]=k++;
}
void dfs(int u,int dep)
{
check[u]=true;
ver[++tot]=u;
first[u]=tot;
depth[tot]=dep;
for(int k=head[u];k!=-1;k=edge[k].next)
{
int v=edge[k].e;
int w=edge[k].w;
if(!check[v])
{
dis[v]=dis[u]+w;
dfs(v,dep+1);
ver[++tot]=u;
depth[tot]=dep;
}
}
}
void ST(int len)
{
for(int i=1;i<=len;i++)
Min[i][0]=i;
for(int i=1;(1<y)
swap(x,y);
if(x==y)
return ver[x];
int temp=RMQ(x,y);
return ver[temp];
}
int main()
{
scanf("%d",&t);
while(t--)
{
tot=0;
dis[1]=0;
memset(check,false,sizeof(check));
memset(head,-1,sizeof(head));
scanf("%d%d",&n,&m);
int num=0;
for(int i=1;i