How far away ？ HDU

How far away ？

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 27977    Accepted Submission(s): 11218

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

 

Recommend

lcy

看样子是不能用floyd

 对于无向图可以随便取一个点当作根节点

先求一遍从根节点到其它节点的距离

然后lca  看代码叭

#include <iostream>
#include <cstdio>
#include <sstream>
#include <cstring>
#include <map>
#include <cctype>
#include <set>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <bitset>
#define rap(i, a, n) for(int i=a; i<=n; i++)
#define rep(i, a, n) for(int i=a; i<n; i++)
#define lap(i, a, n) for(int i=n; i>=a; i--)
#define lep(i, a, n) for(int i=n; i>a; i--)
#define rd(a) scanf("%d", &a)
#define rlld(a) scanf("%lld", &a)
#define rc(a) scanf("%c", &a)
#define rs(a) scanf("%s", a)
#define rb(a) scanf("%lf", &a)
#define rf(a) scanf("%f", &a)
#define pd(a) printf("%d
", a)
#define plld(a) printf("%lld
", a)
#define pc(a) printf("%c
", a)
#define ps(a) printf("%s
", a)
#define MOD 2018
#define LL long long
#define ULL unsigned long long
#define Pair pair<int, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define _  ios_base::sync_with_stdio(0),cin.tie(0)
//freopen("1.txt", "r", stdin);
using namespace std;
const int maxn = 110000, INF = 0x7fffffff;
int n, m, s;
int anc[maxn][20], deep[maxn], vis[maxn];
LL d[maxn];
int head[maxn], cnt;
struct node
{
    int u, v, next, c;
}Node[maxn << 1];

void add_(int u, int v, int c)
{
    Node[cnt].u = u;
    Node[cnt].v = v;
    Node[cnt].c = c;
    Node[cnt].next = head[u];
    head[u] = cnt++;
}

void add(int u, int v, int  c)
{
    add_(u, v, c);
    add_(v, u, c);
}

int dfs(int u, int fa)
{
    for(int i = 1; i < 20; i++)
        anc[u][i] = anc[anc[u][i - 1]][i - 1];
    for(int i = head[u]; i != -1; i = Node[i].next)
    {
        int v = Node[i].v;
        if(v == fa || deep[v]) continue;
        anc[v][0] = u;
        deep[v] = deep[u] + 1;
        dfs(v, u);
    }
}

int lca(int u, int v)
{
    if(deep[u] < deep[v]) swap(u, v);
    for(int i = 20 - 1; i >= 0; i--)
        if(deep[anc[u][i]] >= deep[v])
            u = anc[u][i];

    for(int i = 20 - 1; i >= 0; i--)
    {
        if(anc[u][i] != anc[v][i])
        {
            u = anc[u][i];
            v = anc[v][i];
        }
    }
    if(u == v) return u;
    return anc[u][0];
}



void spfa()
{
    for(int i = 0; i < maxn; i++) d[i] = INF;
   // cout<< d[2] << endl;
    deque<int> Q;
    Q.push_back(s);
    d[s] = 0;
    vis[s] = 1;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop_front();
        vis[u] = 0;
        for(int i = head[u]; i != -1; i = Node[i].next)
        {
            int v = Node[i].v;
           // cout << v << endl;
            if(d[v] > d[u] + Node[i].c)
            {
                //cout << 2222 << endl;
                d[v] = d[u] + Node[i].c;
                if(!vis[v])
                {
                    if(Q.empty()) Q.push_front(v);
                    else if(d[Q.front()] > d[v]) Q.push_front(v);
                    else Q.push_back(v);
                    vis[v] = 1;
                }
            }
        }
    }
}



void init()
{
    mem(head, -1);
    cnt = 0;
    mem(vis, 0);
    mem(anc, 0);
    mem(deep, 0);
}

int main()
{

    int T;
    rd(T);
    while(T--)
    {
        init();
        rd(n), rd(m);
        s = 1;
        for(int i = 1; i < n; i++)
        {
            int u, v, w;
            rd(u), rd(v), rd(w);
            add(u, v, w);
        }
        spfa();
        deep[s] = 1;
        dfs(s, -1);
        for(int i = 0; i < m; i++)
        {
            int u, v;
            rd(u), rd(v);
            int x = lca(u, v);
            printf("%d
", d[u] + d[v] - 2 * d[x]);

        }



    }


    return 0;
}