hdu 2586 How far away ？ (LCA)

http://acm.hdu.edu.cn/showproblem.php?pid=2586

　　题意是给出一棵树，求两点之间的距离。

　　经典LCA的题，在hdu那里无意中看到的。做法是dfs出一个先根遍历的路径，然后对路径求RMQ，当然，也可以用线段树维护。之后就是对每个询问输出dis[u]+dis[v]-dis[lca(u,v)]*2的值即可。

代码如下：

//#pragma comment(linker, "/STACK:102400000,102400000")

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <iostream>

using namespace std;

const int N = 111111;
const int M = 44444;
int rmq[18][N];

void buildRMQ(int n, int *arr) {
    for (int i = 0; i < n; i++) rmq[0][i] = arr[i];
    for (int len = 1; (1 << len) <= n; len++) {
        for (int i = 0; i <= n - len; i++) {
            rmq[len][i] = min(rmq[len - 1][i], rmq[len - 1][i + (1 << len - 1)]);
        }
    }
}

int queryRMQ(int l, int r) {
    if (l > r) swap(l, r);
    int depth;
    for (depth = 0; (1 << depth + 1) <= r - l; depth++) ;
    return min(rmq[depth][l], rmq[depth][r - (1 << depth) + 1]);
}

//int test[10] = { 10, 3, 9, 5, 1, 9, 10, 11, 6, 3};
//int main() {
//    buildRMQ(10, test);
//    for (int i = 0; i <= 4; i++) {
//        for (int j = 0; j < 10; j++) {
//            cout << rmq[i][j] << ' ';
//        }
//        cout << endl;
//    }
//    cout << queryRMQ(2, 5) << endl;
//    cout << queryRMQ(6, 8) << endl;
//    return 0;
//}

#define MPR make_pair
#define PB push_back

typedef pair<int, int> PII;
vector<PII> rel[M];
bool vis[M];
int dis[M], path[N], id[M], pLen;

void dfs(int x) {
    vis[x] = true;
    int sz = rel[x].size();
    path[pLen] = x;
    id[x] = pLen++;
    for (int i = 0, t; i < sz; i++) {
        t = rel[x][i].first;
        if (vis[t]) continue;
        dis[t] = dis[x] + rel[x][i].second;
        dfs(t);
        path[pLen++] = x;
    }
}

int query(int x, int y) {
    int t = queryRMQ(id[x], id[y]);
    return dis[x] - dis[t] + dis[y] - dis[t];
}

int main() {
//    freopen("in", "r", stdin);
    int n, m, T;
    while (~scanf("%d", &T)) {
        while (T-- && ~scanf("%d%d", &n, &m)) {
            for (int i = 1; i <= n; i++) {
                rel[i].clear();
                vis[i] = false;
            }
            int x, y, c;
            for (int i = 1; i < n; i++) {
                scanf("%d%d%d", &x, &y, &c);
//                printf("%d %d %d...\n", x, y, c);
                rel[x].PB(MPR(y, c));
                rel[y].PB(MPR(x, c));
            }
            pLen = 0;
            dis[1] = 0;
            dfs(1);
//            for (int i = 0; i < pLen; i++) cout << path[i] << ' '; cout << endl;
//            for (int i = 0; i < pLen; i++) cout << dis[i] << ' '; cout << endl;
            buildRMQ(pLen, path);
            for (int i = 0; i < m; i++) {
                scanf("%d%d", &x, &y);
                printf("%d\n", query(x, y));
            }
        }
    }
    return 0;
}

——written by Lyon