Component Tree

题意：给定一棵树，每一个节点都有一些属性以及他们的状态，然后给定q个询问，每个询问问节点A的某个属性的状态，如果没有就往上咨询，直到有为止，如果没有就输出“N/A”（题目强制在线操作）。



解法：可持久化线段树，先对属性离散化（我用的是map），线段树维护当前节点以及往上的每一个属性的状态就好了，直接裸地维护就好了。

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

const int maxn = 300100;

map<string, int> g;
string s1, s2;
struct Node {
    int l, r, l_point, r_point;
    string s;
    Node() {
        l = 0; r = 0;
        l_point = 0; r_point = 0;
        s = "";
    }
}a[20*maxn];
char c[30];
int last[maxn], pre[maxn];
int tot, n, m, start, num, syg;

void add_point(int cor_x, int x, int tar, string _s) {
    if (a[x].l == a[x].r) {
        a[x].s = _s;
        return;
    }
    int mid = (a[x].l+a[x].r)/2;
    if (tar <= mid) {
        a[x].l_point = ++tot;
        a[tot].l = a[x].l;
        a[tot].r = mid;
        a[x].r_point = a[cor_x].r_point;
        add_point(a[cor_x].l_point, tot, tar, _s);
    } else {
        a[x].l_point = a[cor_x].l_point;
        a[x].r_point = ++tot;
        a[tot].l = mid+1;
        a[tot].r = a[x].r;
        add_point(a[cor_x].r_point, tot, tar, _s);
    }
}

string find(int x, int tar) {
    if (x == 0) return "N/A";
    if (a[x].l == a[x].r) return a[x].s;
    int mid = (a[x].l+a[x].r)/2;
    if (tar <= mid) {
        return find(a[x].l_point, tar);
    } else {
        return find(a[x].r_point, tar);
    }
}

int main() {
    scanf("%d", &n);
    num = 0;
    tot = 0;
    syg = 0;
    for (int i = 1; i <= n; i++) {
        scanf("%d%d", &pre[i], &m);
        syg = last[pre[i]];
        if (m == 0) last[i] = last[pre[i]];
        while (m--) {
            scanf("%s", c);
            int len = strlen(c);
            int j;
            s1 = "";
            for (j = 0; j < len; j++) {
                if (c[j] == '=') break;
                s1 = s1+c[j];
            }
            if (!g[s1]) g[s1] = ++num;
            s2 = "";
            for (j = j+1; j < len; j++) s2 = s2+c[j];

            last[i] = tot+1;
            int temp = tot+1;
            tot++;
            a[tot].l = 1; a[tot].r = maxn;
            a[tot].l_point = a[tot].r_point = 0;
            a[tot].s = "";
            add_point(syg, tot, g[s1], s2);
            syg = temp;
        }
    }
    scanf("%d", &m);
    while (m--) {
        scanf("%d %s", &start, c);
        s1 = c;
        if (!g[s1]) {
            cout << "N/A" << endl; cout.flush(); continue;
        }
        cout << find(last[start], g[s1]) << endl; cout.flush();
    }
}