Design the city（LCA + RMQ）

Design the city

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Time Limit: 1 Second      Memory Limit: 32768 KB

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Cerror is the mayor of city HangZhou. As you may know, the traffic system of this city is so terrible, that there are traffic jams everywhere. Now, Cerror finds out that the main reason of them is the poor design of the roads distribution, and he want to change this situation.

In order to achieve this project, he divide the city up to N regions which can be viewed as separate points. He thinks that the best design is the one that connect all region with shortest road, and he is asking you to check some of his designs.

Now, he gives you an acyclic graph representing his road design, you need to find out the shortest path to connect some group of three regions.

Input

The input contains multiple test cases! In each case, the first line contian a interger N (1 < N < 50000), indicating the number of regions, which are indexed from 0 to N-1. In each of the following N-1 lines, there are three interger Ai, Bi, Li (1 < Li < 100) indicating there's a road with length Li between region Ai and region Bi. Then an interger Q (1 < Q < 70000), the number of group of regions you need to check. Then in each of the following Q lines, there are three interger Xi, Yi, Zi, indicating the indices of the three regions to be checked.

Process to the end of file.

Output

Q lines for each test case. In each line output an interger indicating the minimum length of path to connect the three regions.

Output a blank line between each test cases.

Sample Input

4
0 1 1
0 2 1
0 3 1
2
1 2 3
0 1 2
5
0 1 1
0 2 1
1 3 1
1 4 1
2
0 1 2
1 0 3

Sample Output

 

3
2

2
2

 

       题意：

       给出结点数 N ，后给出 N - 1 条边。给出 Q 个询问，每个询问给出 3 个结点，输出这三点连通的最短距离。

 

       思路：

       LCA。因为 1 -> 2 + 2 -> 3 的距离等于 1 -> 3 的距离，所以将 3 个点间两两的距离求出后加和除以 2 即为答案。用 LCA 求出两两间点距离即可。输出的格式问题，每个 case 后输出一行空行，最后一组样例不需要有空行。所以用 temp 标记输出即可。

 

       AC：

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

using namespace std;

const int VMAX = 50010;
const int EMAX = VMAX * 2;

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

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

int dp[VMAX * 2][25];

void init () {
    ind = cnt = 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) {
    id[x] = cnt;
    vs[cnt] = x;
    dep[cnt++] = d;
    vis[x] = 1;

    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);
            vs[cnt] = x;
            dep[cnt++] = d;
        }
    }
}

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

    for (int j = 1; (1 << j) <= cnt; ++j) {
        for (int i = 0; i + (1 << j) < cnt; ++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 << (1 + len)) <= 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 Distance (int a, int b, int c) {
    int ab = LCA(a, b);
    int ac = LCA(a, c);
    int bc = LCA(b, c);
    int res = 0;
    res += dis[a] + dis[b] - 2 * dis[ab];
    res += dis[a] + dis[c] - 2 * dis[ac];
    res += dis[b] + dis[c] - 2 * dis[bc];
    return res / 2;
}

int main() {

    int temp = 0;

    while (~scanf("%d", &n)) {
        if (temp) printf("\n");

        temp = 1;
        init();

        for (int i = 1; i <= n - 1; ++i) {
            int a, b, val;
            scanf("%d%d%d", &a, &b, &val);
            ++a; ++b;
            add_edge(a, b, val);
            add_edge(b, a, val);
        }

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

        int q;
        scanf("%d", &q);
        while (q--) {
            int a, b, c;
            scanf("%d%d%d", &a, &b, &c);
            ++a, ++b, ++c;
            printf("%d\n", Distance(a, b, c));
        }

    }

    return 0;
}