hdu 2586 How far away ？ (LCA转RMQ)

题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=2586

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 18268    Accepted Submission(s): 7106

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

23 21 2 103 1 151 22 32 21 2 1001 22 1

 

Sample Output

1025100100

 

Source

ECJTU 2009 Spring Contest

 

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解析：裸的模板Lca，距离[u->v] = dis[u] + dis[v] - 2*dis[lca]; dis[k]代表节点k到树根的距离

代码：

#include<bits/stdc++.h>using namespace std;struct node{    int to, val;};vector<node> mp[40001];int n, m;int fr[40001], dep[40001], dis[40001];int zx, vis[40001], visnum[80001];int dp[80001][20];void creat(int u, int k, int d){    dep[u] = k;    dis[u] = d;    for(int i = 0; i < mp[u].size(); i++)    {        node cur = mp[u][i];        vis[u] = zx;        visnum[zx++] = u;        creat(cur.to, k+1, d+cur.val);    }    vis[u] = zx;    visnum[zx++] = u;}void rmq(){    int s = n*2-1;    for(int i = 1; i <= s; i++) dp[i][0] = visnum[i];    for(int j = 1; j != 20; j++)    {        for(int i = 1; i <= s; i++)        {            if(i + (1<<j) - 1 > s) continue;            if(dep[dp[i][j - 1]] < dep[dp[i + (1<<j>>1)][j - 1]]) dp[i][j] = dp[i][j - 1];            else dp[i][j] = dp[i + (1<<j>>1)][j - 1];        }    }}int lca(int x, int y){    int k = 0;    while((1<<k) < y-x+1) k++;    k--;    int ans;    if(dep[dp[y - (1<<k)+1][k]] < dep[dp[x][k]]) ans = dp[y - (1<<k)+1][k];    else ans = dp[x][k];    return ans;}int main(){    int t;    node cur;    scanf("%d", &t);    while(t--)    {        int u, v, w;        scanf("%d%d", &n, &m);        for(int i = 1; i <= n; i++) mp[i].clear();        for(int i = 1; i <= n; i++) fr[i] = i;        for(int i = 1; i < n; i++)        {            scanf("%d%d%d", &u, &v, &w);            cur.to = v; cur.val = w;            mp[u].push_back(cur);            fr[v] = u;        }        while(fr[u] != u) u = fr[u];        zx = 1;        creat(u, 1, 0);        rmq();        while(m--)        {            scanf("%d%d", &u, &v);            int lca_;            if(vis[u] < vis[v]) lca_ = lca(vis[u], vis[v]);            else lca_ = lca(vis[v], vis[u]);            printf("%d\n", dis[u] + dis[v] - 2*dis[lca_]);        }    }    return 0;}