Distance Queries（LCA 在线算法 RMQ）

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本文深入探讨了深度学习在人工智能领域的应用，包括卷积神经网络、循环神经网络、自动推理系统等关键技术，并展示了这些技术在自动驾驶、图像处理、自然语言处理等实际场景中的应用案例。

Distance Queries

Time Limit: 2000MS		Memory Limit: 30000K
Total Submissions: 9500		Accepted: 3332
Case Time Limit: 1000MS

Description

Farmer John's cows refused to run in his marathon since he chose a path much too long for their leisurely lifestyle. He therefore wants to find a path of a more reasonable length. The input to this problem consists of the same input as in "Navigation Nightmare",followed by a line containing a single integer K, followed by K "distance queries". Each distance query is a line of input containing two integers, giving the numbers of two farms between which FJ is interested in computing distance (measured in the length of the roads along the path between the two farms). Please answer FJ's distance queries as quickly as possible!

Input

* Lines 1..1+M: Same format as "Navigation Nightmare"

* Line 2+M: A single integer, K. 1 <= K <= 10,000

* Lines 3+M..2+M+K: Each line corresponds to a distance query and contains the indices of two farms.

Output

* Lines 1..K: For each distance query, output on a single line an integer giving the appropriate distance.

Sample Input

Sample Output

13
3
36

Hint

Farms 2 and 6 are 20+3+13=36 apart.

题意：

给出 N 和 M，代表 N 个节点，M 条边，后给出边信息，有 K 个询问，输出每个询问两点间的距离。

思路：

LCA。RMQ 的在线算法。

AC：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int VMAX = 40010;
const int EMAX = VMAX * 5;

int ind;
int v[EMAX], fir[VMAX], next[EMAX], w[EMAX];

int ans;
int id[VMAX], vs[VMAX * 2], dep[VMAX * 2], dis[VMAX];
bool vis[VMAX];

int dp[VMAX * 2][30];

void init () {
    ind = ans = 0;
    memset(fir, -1, sizeof(fir));
    memset(vis, 0, sizeof(vis));
}

void add_edge (int f, int t, int val) {
    v[ind] = t;
    w[ind] = val;
    next[ind] = fir[f];
    fir[f] = ind;
    ++ind;
}

void dfs (int x, int d) {
    vis[x] = 1;
    id[x] = ans;
    dep[ans] = d;
    vs[ans++] = x;

    for (int e = fir[x]; e != -1; e = next[e]) {
        int V = v[e];
        if (!vis[V]) {
            dis[V] = dis[x] + w[e];
            dfs(V, d + 1);
            dep[ans] = d;
            vs[ans++] = x;
        }
    }
}

void RMQ_init () {
    for (int i = 0; i < ans; ++i) dp[i][0] = i;

    for (int j = 1; (1 << j) <= ans; ++j) {
        for (int i = 0; i + (1 << j) < ans; ++i) {
            int a = dp[i][j - 1];
            int b = dp[i + (1 << (j - 1))][j - 1];
            if (dep[a] < dep[b]) dp[i][j] = a;
            else dp[i][j] = b;
        }
    }
}

int RMQ (int L, int R) {
    int len = 0;
    while ((1 << (len + 1)) <= (R - L + 1)) ++len;

    int a = dp[L][len];
    int b = dp[R - (1 << len) + 1][len];

    if (dep[a] < dep[b]) return a;
    return b;
}

int LCA (int a, int b) {
    int L = min(id[a], id[b]);
    int R = max(id[a], id[b]);

    int node = RMQ(L, R);
    return vs[node];
}

int main () {

    init();

    int n, m;
    scanf("%d%d", &n, &m);

    while (m--) {
        int a, b, val;
        char c;
        scanf("%d%d%d %c", &a, &b, &val, &c);
        add_edge(a, b, val);
        add_edge(b, a, val);
    }

    dis[1] = 0;
    dfs(1, 1);

    RMQ_init();

    int k;
    scanf("%d", &k);
    while (k--) {
        int a, b;
        scanf("%d%d", &a, &b);

        int c = LCA(a, b);
        printf("%d\n", dis[a] + dis[b] - 2 * dis[c]);
    }

    return 0;
}