hdu2586 LCA

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.

 

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 = 40010;
struct node{
    int to;
    int val;
    int next;
}edge[MAXN*2];
int dp[MAXN*2][20],n,m,ind,pre[MAXN],vis[MAXN];
int dfn[MAXN*2],first[MAXN],ver[MAXN*2],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] = 1;
    ver[++tot] = rt;
    dfn[tot] = dep;
    first[rt] = tot;
    for(int i = pre[rt]; i != -1; i = edge[i].next){
        int t = edge[i].to;
        if(!vis[t]){
            dis[t] = dis[rt] + edge[i].val;
            dfs(t,dep+1);
            ver[++tot] = rt;//先序访问
            dfn[tot] = dep;
        }
    }
}
void rmq(){
    for(int i = 1; i <= tot; i++){
        dp[i][0] = i;
    }
    for(int i = 1; i < 20; i++){
        for(int j = 1; j + (1 << i) - 1 <= tot; j++){
            if(dfn[dp[j][i-1]] > dfn[dp[j+(1<<(i-1))][i-1]]){
                dp[j][i] = dp[j+(1<<(i-1))][i-1];
            } else {
                dp[j][i] = dp[j][i-1];
            }
        }
    }
}
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) * 1.0 / log(2.0));
    int l = dp[x][k];
    int r = dp[y-(1<<k)+1][k];
    if(dfn[l] > dfn[r])return r;
    else return l;
}
int main(){
    int t,flag = 0;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        ind = 0;
        tot = 0;
        memset(vis,0,sizeof(vis));
        memset(dis,0,sizeof(dis));
        memset(pre,-1,sizeof(pre));
        for(int i = 1; i < n; i++){
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            add(x,y,z),add(y,x,z);
        }
        dfs(1,1);
        rmq();
        if(flag)puts("");
        flag = 1;
        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
",ans);
        }
    }
    return 0;
}