[BZOJ4771]七彩树

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题目链接：BZOJ4771

题目大意
有一棵染色的树，每个点有一个颜色，若干个询问， Q　x　d，询问x的子树里与x的深度差不超过d的点集有多少个不同的颜色。

分析
1. 建两棵主席树，T1和T2；
2. T1[x]维护x子树中在l到r的深度区间内，每个区间有多少个不同的颜色，若某一个颜色同时出现在两个深度内，则只保存深度较小的那一个。
3. T2[x]维护x子树中在l到r的颜色区间内，每个颜色出现的最浅深度；
4. 自底向上合并T1，T2；当合并T2时，发现有某个颜色出现在deepi和deepj(deepi≤deepj)里面，则在T1的[deepj,deepj]区间-1。
5. 询问时直接在T1[x]中查询就好了，时空复杂度O(NlogN)。

上代码

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N = 1e5 + 10;const int M = 1e7 + 10;int n, m;inline int read() {    char ch;    int ans = 0, neg = 1;    while (!isdigit(ch = getchar()))        if (ch == '-') neg = -1;    while (isdigit(ch))        ans = ans * 10 + ch - '0', ch = getchar();    return ans * neg;}int fa[N], clr[N], dep[N];int T1[N], T2[N], cnt;int lc[M], rc[M], tot[M];int modify(int a, int l, int r, int p, int c) {    int now = ++cnt;    tot[now] = tot[a] + c;    if (l == r) return now;    int mid = (l + r) >> 1;    if (p <= mid) lc[now] = modify(lc[a], l, mid, p, c), rc[now] = rc[a];    else rc[now] = modify(rc[a], mid + 1, r, p, c), lc[now] = lc[a];    return now;}int merge1(int a, int b, int l, int r) {    if (!a) return b;    if (!b) return a;    int now = ++cnt;    tot[now] = tot[a] + tot[b];    if (l == r) return now;    int mid = (l + r) >> 1;    lc[now] = merge1(lc[a], lc[b], l, mid);    rc[now] = merge1(rc[a], rc[b], mid + 1, r);    return now;}int merge2(int a, int b, int l, int r, int p) {    if (!a) return b;    if (!b) return a;    int now = ++cnt;    if (l == r) {        if (tot[a] > tot[b]) swap(a, b);        tot[now] = tot[a], T1[p] = modify(T1[p], 1, n, tot[b], -1);        return now;    }    int mid = (l + r) >> 1;    lc[now] = merge2(lc[a], lc[b], l, mid, p);    rc[now] = merge2(rc[a], rc[b], mid + 1, r, p);    return now;}int query(int a, int l, int r, int ll, int rr) {    if (l == ll && r == rr) return tot[a];    int mid = (l + r) >> 1;    if (rr <= mid) return query(lc[a], l, mid, ll, rr);    else if (ll > mid) return query(rc[a], mid + 1, r, ll, rr);    return query(lc[a], l, mid, ll, mid) + query(rc[a], mid + 1, r, mid + 1, rr);}void init() {    n = read(), m = read(), cnt = 0;    for (int i = 1; i <= n; i++) clr[i] = read();    for (int i = 2; i <= n; i++) fa[i] = read();    for (int i = 1; i <= n; i++) dep[i] = dep[fa[i]] + 1;    for (int i = 1; i <= n; i++) {        T1[i] = modify(0, 1, n, dep[i], 1);        T2[i] = modify(0, 1, n, clr[i], dep[i]);    }    for (int i = n; i > 1; i--) {        T1[fa[i]] = merge1(T1[fa[i]], T1[i], 1, n);        T2[fa[i]] = merge2(T2[fa[i]], T2[i], 1, n, fa[i]);    }}void figure() {    int ans = 0;    for (int i = 1; i <= m; i++) {        int a = read() ^ ans, b = read() ^ ans;        ans = query(T1[a], 1, n, dep[a], min(n, dep[a] + b));        printf("%d\n", ans);    }}int main() {    int T = read();    while (T--) {        init();        figure();    }    return 0;}

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