How far away ？
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11739 Accepted Submission(s): 4325
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.
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:
n个点组成的一棵树，2点之间有长度。m次查询，问x到y上每个点只能走一次的最近的距离是多少。
思路:
是一棵树，又x到y的路径上每个点只能走一次，dfs时记录当前节点到根的距离，然后对于查询x,y,求出lca(x,y),答案就是dis[x] + dis[y] - 2 * dis[lca(x,y)];
/*
* Author: sweat123
* Created Time: 2016/7/13 8:53:48
* File Name: main.cpp
*/
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<vector>
#include<cstdio>
#include<time.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1<<30
#define MOD 1000000007
#define ll long long
#define lson l,m,rt<<1
#define key_value ch[ch[root][1]][0]
#define rson m+1,r,rt<<1|1
#define pi acos(-1.0)
using namespace std;
const int MAXN = ;
struct node{
int to;
int val;
int next;
}edge[MAXN*];
int dp[MAXN*][],n,m,ind,pre[MAXN],vis[MAXN];
int dfn[MAXN*],first[MAXN],ver[MAXN*],tot;
ll dis[MAXN];
//dfn 表示深度 first第一次出现这个点的下标 dis长度 ver先序访问的节点编号 tot编号个数
void add(int x,int y,int z){
edge[ind].to = y;
edge[ind].val = z;
edge[ind].next = pre[x];
pre[x] = ind ++;
}
void dfs(int rt,int dep){
vis[rt] = ;
ver[++tot] = rt;
dfn[tot] = dep;
first[rt] = tot;
for(int i = pre[rt]; i != -; i = edge[i].next){
int t = edge[i].to;
if(!vis[t]){
dis[t] = dis[rt] + edge[i].val;
dfs(t,dep+);
ver[++tot] = rt;//先序访问
dfn[tot] = dep;
}
}
}
void rmq(){
for(int i = ; i <= tot; i++){
dp[i][] = i;
}
for(int i = ; i < ; i++){
for(int j = ; j + ( << i) - <= tot; j++){
if(dfn[dp[j][i-]] > dfn[dp[j+(<<(i-))][i-]]){
dp[j][i] = dp[j+(<<(i-))][i-];
} else {
dp[j][i] = dp[j][i-];
}
}
}
}
int askrmq(int x,int y){
x = first[x];
y = first[y];
if(x > y)swap(x,y);
int k = (int)(log(y - x + ) * 1.0 / log(2.0));
int l = dp[x][k];
int r = dp[y-(<<k)+][k];
if(dfn[l] > dfn[r])return r;
else return l;
}
int main(){
int t,flag = ;
scanf("%d",&t);
while(t--){
scanf("%d%d",&n,&m);
ind = ;
tot = ;
memset(vis,,sizeof(vis));
memset(dis,,sizeof(dis));
memset(pre,-,sizeof(pre));
for(int i = ; i < n; i++){
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
add(x,y,z),add(y,x,z);
}
dfs(,);
rmq();
if(flag)puts("");
flag = ;
while(m--){
int x,y;
scanf("%d%d",&x,&y);
int tp = ver[askrmq(x,y)];
ll ans = dis[x] - dis[tp] + dis[y] - dis[tp];
printf("%lld\n",ans);
}
}
return ;
}