Codeforces379F-New Year Tree（LCA）

题目链接

http://codeforces.com/problemset/problem/379/F

思路

每次维护当前的直径d和直径的两个端点u，v。当新加入两个点x和y时，选择x或者y相同，因此只考虑x：
当x到u的距离大于原来的直径时，将v更新为x，并更新直径
当x到v的距离大于原来的直径时，将u更新为x，并更新直径
每次使用LCA在线维护直径

代码

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

#define PI acos(-1.0)
#define LL long long
#define PII pair
#define PLL pair
#define mp make_pair
#define IN freopen("in.txt", "r", stdin)
#define OUT freopen("out.txt", "wb", stdout)
#define scan(x) scanf("%d", &x)
#define scan2(x, y) scanf("%d%d", &x, &y)
#define scan3(x, y, z) scanf("%d%d%d", &x, &y, &z)
#define sqr(x) (x) * (x)

const int maxn = 1000000 + 5;
int p[maxn][32], dep[maxn];
int Q, tot = 4;

int getDist(int u, int v) {
    if (dep[u] < dep[v]) swap(u, v);
    int tu = u, tv = v;
    for (int i = 20; i >= 0; i--) {
        if (dep[u] - (1 << i) < dep[v]) continue;
        u = p[u][i];
    }
    if (u != v) {
        for (int i = 20; i >= 0; i--) {
            if (p[u][i] == -1 || p[v][i] == -1 || p[u][i] == p[v][i]) continue;
                u = p[u][i];
                v = p[v][i];
        }
    }
    u = p[u][0];
    return dep[tu] + dep[tv] - 2 * dep[u];
}

int main() {
    dep[1] = 1; dep[2] = dep[3] = dep[4] = 2;
    p[1][0] = -1;
    p[2][0] = p[3][0] = p[4][0] = 1;
    scan(Q);
    int x, u = 2, v = 3, d = 2;
    while (Q--) {
        scan(x);
        p[++tot][0] = x; dep[tot] = dep[x] + 1;
        p[++tot][0] = x; dep[tot] = dep[x] + 1;
        for (int j = 1; (1 << j) <= tot; j++) {
            if (p[tot - 1][j - 1] != -1) p[tot - 1][j] = p[p[tot - 1][j - 1]][j - 1];
            if (p[tot - 2][j - 1] != -1) p[tot - 2][j] = p[p[tot - 2][j - 1]][j - 1];
        }
        int d1 = getDist(tot - 1, u);
        int d2 = getDist(tot - 1, v);
        if (d1 > d) {
            v = tot - 1;
            d = d1;
        }
        if (d2 > d) {
            u = tot - 1;
            d = d2;
        }
        printf("%d\n", d);
    }
    return 0; 
}