在线lca算法模板

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

题意：求树上两点间距离

#include 
#include 
#include 
#include 
#include 
using namespace std;

const int N = 100100;
struct edge
{
    int to, cost, next;
}g[N*2];

int dp[20][N*2];
//dp[i][j]记录的是以j为起点、2的i次方长度内的深度最小点的编号，因为访问2n-1次，数组长度要乘2
int n, m;
int cnt, tot, head[N];
bool vis[N];
int dep[N*2], ver[N*2], fir[N], dis[N]; 
//ver为访问的节点依次编号，dep记录编号对应的节点的深度，因为访问2n-1次，所有这两个数组长度乘2

void add_edge(int v, int u, int cost)
{
    g[cnt].to = u, g[cnt].cost = cost, g[cnt].next = head[v], head[v] = cnt++;
}
void dfs(int v, int cur)
{
    vis[v] = true, ver[++tot] = v, dep[tot] = cur, fir[v] = tot;
    for(int i = head[v]; i != -1; i = g[i].next)
    {
        int u = g[i].to;
        if(! vis[u])
        {
            dis[u] = dis[v] + g[i].cost;
            dfs(u, cur + 1);
            ver[++tot] = v, dep[tot] = cur;
        }
    }
}
void ST(int n)
{
    for(int i = 1; i <= n; i++)
        dp[0][i] = i;
    for(int i = 1; (1< u) swap(v, u);
    int res = rmq(v, u);
    return ver[res];
}
int main()
{
    int t, a, b, c;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d%d", &n, &m);
        cnt = 0;
        memset(head, -1, sizeof head);
        for(int i = 0; i < n - 1; i++)
        {
            scanf("%d%d%d", &a, &b, &c);
            add_edge(a, b, c);
            add_edge(b, a, c);
        }
        memset(vis, 0, sizeof vis);
        dis[1] = 0;
        tot = 0;
        dfs(1, 1);
        ST(2 * n - 1);
        for(int i = 0; i < m; i++)
        {
            scanf("%d%d", &a, &b);
            int res = lca(a, b);
            printf("%d\n", dis[a] + dis[b] - 2 * dis[res]);
        }
    }
    return 0;
}