How far away ？ LCA求树上两点距离

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22976    Accepted Submission(s): 9078

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

 

 

Source

ECJTU 2009 Spring Contest

 

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
#include <algorithm>
#include <set>
#include <iostream>
#include <map>
#include <stack>
#include <string>
#include <vector>
#define  pi acos(-1.0)
#define  eps 1e-6
#define  fi first
#define  se second
#define  lson l,m,rt<<1
#define  rson m+1,r,rt<<1|1
#define  bug         printf("******\n")
#define  mem(a,b)    memset(a,b,sizeof(a))
#define  fuck(x)     cout<<"["<<x<<"]"<<endl
#define  f(a)        a*a
#define  sf(n)       scanf("%d", &n)
#define  sff(a,b)    scanf("%d %d", &a, &b)
#define  sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define  sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d)
#define  pf          printf
#define  FRE(i,a,b)  for(i = a; i <= b; i++)
#define  FREE(i,a,b) for(i = a; i >= b; i--)
#define  FRL(i,a,b)  for(i = a; i < b; i++)
#define  FRLL(i,a,b) for(i = a; i > b; i--)
#define  FIN         freopen("DATA.txt","r",stdin)
#define  gcd(a,b)    __gcd(a,b)
#define  lowbit(x)   x&-x
#pragma  comment (linker,"/STACK:102400000,102400000")
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1e5 + 10;
int _pow[maxn], dep[maxn], dis[maxn], vis[maxn], ver[maxn];
int tot, head[maxn], dp[maxn * 2][25], k, first[maxn];
struct node {
    int u, v, w, nxt;
} edge[maxn << 2];
void init() {
    tot = 0;
    mem(head, -1);
}
void add(int u, int v, int w) {
    edge[tot].v = v, edge[tot].u = u;
    edge[tot].w = w, edge[tot].nxt = head[u];
    head[u] = tot++;
}
void dfs(int u, int DEP) {
    vis[u] = 1;
    ver[++k] = u;
    first[u] = k;
    dep[k] = DEP;
    for (int i = head[u]; ~i; i = edge[i].nxt) {
        if (vis[edge[i].v]) continue;
        int v = edge[i].v, w = edge[i].w;
        dis[v] = dis[u] + w;
        dfs(v, DEP + 1);
        ver[++k] = u;
        dep[k] = DEP;
    }
}
void ST(int len) {
    int K = (int)(log((double)len) / log(2.0));
    for (int i = 1 ; i <= len ; i++) dp[i][0] = i;
    for (int j = 1 ; j <= K ; j++) {
        for (int i = 1 ; i + _pow[j] - 1 <= len ; i++) {
            int a = dp[i][j - 1], b = dp[i + _pow[j - 1]][j - 1];
            if (dep[a] < dep[b]) dp[i][j] = a;
            else dp[i][j] = b;
        }
    }
}
int RMQ(int x, int y) {
    int K = (int)(log((double)(y - x + 1)) / log(2.0));
    int a = dp[x][K], b = dp[y - _pow[K] + 1][K];
    if (dep[a] < dep[b]) return a;
    else return b;
}
int LCA(int u, int v) {
    int x = first[u], y = first[v];
    if (x > y) swap(x, y);
    int ret = RMQ(x, y);
    return ver[ret];
}
int main() {
    for (int i = 0 ; i < 40 ; i++) _pow[i] = (1 << i);
    int t;
    sf(t);
    while(t--) {
        int n, q;
        init();
        sff(n, q);
        mem(vis, 0);
        for (int i = 1 ; i < n ; i++) {
            int u, v, w;
            sfff(u, v, w);
            add(u, v, w);
            add(v, u, w);
        }
        k = 0, dis[1] = 0;
        dfs(1, 1);
        ST(2 * n - 1);
        while(q--) {
            int u, v;
            sff(u, v);
            int lca = LCA(u, v);
            printf("%d\n", dis[u] + dis[v] - 2 * dis[lca]);
        }
    }
    return  0;
}