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LCA最近公共祖先-- HDU 2586

2019年05月06日  | 移动技术网IT编程  | 我要评论

周觅骂韩庚,测公司名,3d预测号码

 

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 by 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 are to 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：有一棵有n个节点的树，每条边上有一个权值代表这两个点之间的距离，现在m次询问：从节点a到节点b的路径长？

思路：预处理所有节点到根节点（定为节点1）的距离，以及所有节点的祖先信息（fa[i][j]表示节点 i 向上距离为 （1<<j）的祖先节点编号），计算a和b到根节点的距离和，减去两倍的最近公共祖先的到根节点的距离值。

代码如下：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int n = 4e4 + 5;

int head[n], cnt;
struct edge
{
    int to, next;
    int value;
}e[2 * n];

struct node {
    int fa[20];
    int deep;
    int sum;
    bool state;
}node[n];

void insert(int u, int v, int value)
{
    e[++cnt].to = v;
    e[cnt].next = head[u];
    e[cnt].value = value;
    head[u] = cnt;
    e[++cnt].to = u;
    e[cnt].next = head[v];
    e[cnt].value = value;
    head[v] = cnt;
}
int cal(int x, int t)
{
    for (int i = 0; i <= 19; i++)
        if (t&(1 << i)) x = node[x].fa[i];
    return x;
}
void dfs(int x)
{
    node[x].state = 1;
    for (int i = 1; i <= 19; i++)
    {
        if (node[x].deep<(1 << i))break;
        node[x].fa[i] = node[node[x].fa[i - 1]].fa[i-1];///倍增处理祖先信息
    }
    for (int i = head[x]; i; i = e[i].next)
    {
        if (node[e[i].to].state) continue;
        node[e[i].to].deep = node[x].deep+ 1;
        node[e[i].to].fa[0] = x;
        node[e[i].to].sum = node[x].sum+e[i].value;
        dfs(e[i].to);
    }
}
int lca(int x, int y)///求lca
{
    if (node[x].deep<node[y].deep) swap(x, y);
    x = cal(x, node[x].deep - node[y].deep);
    for (int i = 19; i >= 0; i--)
        if (node[x].fa[i] != node[y].fa[i])
        {
            x = node[x].fa[i];
            y = node[y].fa[i];
        }
    if (x == y)return x;
    else return node[x].fa[0];
}

void init()
{
    cnt = 0;
    memset(head, 0, sizeof(head));
    memset(node, 0, sizeof(node));
}

int main()
{
    int t; cin >> t;
    while (t--)
    {
        init();
        int n, m; cin >> n >> m;
        for (int i = 0; i < n-1; i++) {
            int x, y, v; scanf("%d%d%d",&x,&y,&v);
            insert(x,y,v);
        }
        dfs(1);
        for (int i = 0; i < m; i++) {
            int x, y; scanf("%d%d",&x,&y);
            int pa = lca(x, y);
            int ans = node[x].sum - node[pa].sum + node[y].sum - node[pa].sum;
            cout << ans << endl;
        }
    }
    return 0;
}

 

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