poj1330 LCA离线算法

模版参考：http://blog.csdn.net/non_cease/article/details/7426395

题目：给定一棵树，求两个结点的最近公共祖先。（最基础的LCA问题）

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 10006;
int dp[maxn][15], father[maxn], dep[maxn];
bool hash[maxn], mark[maxn];
int n;

void getDepth(int u) { //从叶节点向根求结点深度
     int v = u, l = 0;

     while (dep[v] < 0) {
          v = father[v];
          l++;
     }
     while (v != u) {
          dep[u] = l + dep[v];
          u = father[u];
          l--;
     }
}
//模版
void DP() {
    int i, j;

    memset (dp, -1, sizeof (dp));
    for (i = 1; i <= n; i++)
        dp[i][0] = father[i];
    for (j = 1; (1<<j) <= n; j++)
        for (i = 1; i <= n; i++)
            //if (dp[i][j-1] != -1)
               dp[i][j] = dp[dp[i][j-1]][j-1];
}

int get_nearest_ancestor(int u, int v) {
    int tmp, log, i;

    if (dep[u] < dep[v]) {
        tmp = u; u = v; v = tmp;
    }
    for (log = 1; (1<<log) <= dep[u]; log++);
    log--;
    for (i = log; i >= 0; i--)
        if (dep[u]-(1<<i) >= dep[v])
            u = dp[u][i];
    if (u == v) return u;
    for (i = log; i >= 0; i--)
        if (dp[u][i] != -1 && dp[u][i] != dp[v][i])
            u = dp[u][i],v = dp[v][i];
    return father[u];
}

int main()
{
    int t, i, x, y;

    scanf ("%d", &t);
    while(t--) {
        memset(hash, false, sizeof (hash));
        memset(mark, false, sizeof (mark));
        memset(dep, -1, sizeof (dep));
        scanf ("%d", &n);
        for (i = 1; i < n; i++) {
            scanf ("%d%d", &x, &y);
            father[y] = x;
            hash[y] = true;
            mark[x] = true;
        }
        for (i = 1; i <= n; i++)
            if (!hash[i]) break;
        father[i] = i;
        dep[i] = 0;
        for (i = 1; i <= n; i++) //遍历所有叶节点求结点的深度
            if (!mark[i])
               getDepth(i);
        DP();
        scanf ("%d %d", &x, &y);
        printf("%d\n", get_nearest_ancestor(x, y));
    }
    return 0;
}