How far away ？ HDU - 2586

时间：2021-08-17 06:01:03

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
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\n", a)

#define plld(a) printf("%lld\n", a)

#define pc(a) printf("%c\n", a)

#define ps(a) printf("%s\n", 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 = , INF = 0x7fffffff;

int n, m, s;

int anc[maxn][], deep[maxn], vis[maxn];

LL d[maxn];

int head[maxn], cnt;

struct node

{

    int u, v, next, c;

}Node[maxn << ];

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 = ; i < ; i++)

        anc[u][i] = anc[anc[u][i - ]][i - ];

    for(int i = head[u]; i != -; i = Node[i].next)

    {

        int v = Node[i].v;

        if(v == fa || deep[v]) continue;

        anc[v][] = u;

        deep[v] = deep[u] + ;

        dfs(v, u);

    }

}

int lca(int u, int v)

{

    if(deep[u] < deep[v]) swap(u, v);

    for(int i =  - ; i >= ; i--)

        if(deep[anc[u][i]] >= deep[v])

            u = anc[u][i];

    for(int i =  - ; i >= ; i--)

    {

        if(anc[u][i] != anc[v][i])

        {

            u = anc[u][i];

            v = anc[v][i];

        }

    }

    if(u == v) return u;

    return anc[u][];

}

void spfa()

{

    for(int i = ; i < maxn; i++) d[i] = INF;

   // cout<< d[2] << endl;

    deque<int> Q;

    Q.push_back(s);

    d[s] = ;

    vis[s] = ;

    while(!Q.empty())

    {

        int u = Q.front(); Q.pop_front();

        vis[u] = ;

        for(int i = head[u]; i != -; 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] = ;

                }

            }

        }

    }

}

void init()

{

    mem(head, -);

    cnt = ;

    mem(vis, );

    mem(anc, );

    mem(deep, );

}

int main()

{

    int T;

    rd(T);

    while(T--)

    {

        init();

        rd(n), rd(m);

        s = ;

        for(int i = ; i < n; i++)

        {

            int u, v, w;

            rd(u), rd(v), rd(w);

            add(u, v, w);

        }

        spfa();

        deep[s] = ;

        dfs(s, -);

        for(int i = ; i < m; i++)

        {

            int u, v;

            rd(u), rd(v);

            int x = lca(u, v);

            printf("%d\n", d[u] + d[v] -  * d[x]);

        }

    }

    return ;

}

相关文章