hdu2586【How far away ？】

[tarjian离线求lca]

hdu2586【How far away ？】

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

Total Submission(s): 13605 Accepted Submission(s): 5101

【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 Input23 21 2 103 1 151 22 32 21 2 1001 22 1Sample Output1025100100Source ECJTU 2009 Spring Contest 

【题目大意】：就是tarjian离线求LCA的裸体。。 但是这道题我卡了很久，，原因就是罪恶的HDU，，，他的M不是200，，，只是不止200，我把M也开成40000才过，，，，事实证明，用OJ必须随意开数组，，，kao.

友链：[这个lca讲的不错](http://blog.csdn.net/hnust_xiehonghao/article/details/9109295)

【代码】

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
using namespace std;
#define cle(x) memset(x,0,sizeof(x))
const int N=40010,M=210;
struct ii{
 int to,ne,v;
 ii(int to=0,int ne=0,int v=0):to(to),ne(ne),v(v){ }
}ed[N*2];
struct tt{
 int to,ne;
 tt(int to =0,int ne=0):to(to),ne(ne){ }
}qu[N*2];
int head[N],first[N],num[N],fa[N],ufa[N];
bool vis[N];
inline int readin(){
 int res=0;
 char ch;
 while((ch = getchar())>'9'||ch<'0');
 res=ch-'0';
 while((ch = getchar())>='0'&&ch<='9')
  res=res*10+ch-'0';
 return res;
}
int lca[M],lca_dis[M],dis[N];
int find_fa(int x){return (x==ufa[x])?x:ufa[x]=find_fa(ufa[x]);}//查找的是ufa,,不是fa,,!!!
void tarjian(int u,int w){
 vis[u]=true;
 dis[u]=w;
 ufa[u]=u;
 for(int i = head[u];i;i=ed[i].ne){
  int v=ed[i].to;
  if(fa[u]==v)continue;
  fa[v]=u;
  tarjian(v,w+ed[i].v);
  ufa[v]=u;
 }
 for(int i = first[u];i;i=qu[i].ne){
  int v=qu[i].to;
  if(!vis[v])continue;
  int p=num[i];
  lca[p]=find_fa(v);
  lca_dis[p]=dis[u]+dis[v]-dis[lca[p]]*2;
 }
}
int main(){
 freopen("hdu.in","r",stdin);
 freopen("hdu.out","w",stdout);
 int T=readin();
 for(int o = 1; o<= T; o++){
  int n=readin(),m=readin();
  cle(ed),cle(qu),cle(head),cle(first),cle(num),cle(lca),cle(lca_dis),cle(fa),cle(ufa);
  cle(dis),cle(vis);//数组记得清空完全。。
  for(int i=1;i<=n-1;i++){
   int x=readin(),y=readin(),z=readin();
   ed[i*2-1]=ii(y,head[x],z);
   head[x]=i*2-1;
   ed[i*2]=ii(x,head[y],z);
   head[y]=i*2;
  }
  for(int i=1; i<=m;i++){
   int x=readin(),y=readin();
   qu[i*2-1]=tt(y,first[x]);
   first[x]=i*2-1;
   qu[i*2]=tt(x,first[y]);
   first[y]=i*2;
   num[i*2-1]=num[i*2]=i;//存储是第几个询问，必须双向。
  }
  fa[1]=1;//这个并不知道是否有问题，，，记得实验一次！！！！
  tarjian(1,0);
  for(int i=1;i<=m;i++)
   printf("%d\n",lca_dis[i]);
 }
}