思路:直接树链剖分,用线段树维护即可,算是树剖的经典题目吧。
代码:
#include <bits/stdc++.h>
#define ls(x) (x << 1)
#define rs(x) ((x << 1) | 1)
using namespace std;
const int maxn = 100010;
int head[maxn], Next[maxn * 2], ver[maxn * 2];
int sz[maxn], son[maxn], d[maxn], dfn[maxn], top[maxn], f[maxn];
int tot, cnt;
int n;
struct SegmentTree {
int val, lz;
int l, r;
};
SegmentTree tr[maxn * 4];
void add(int x, int y) {
ver[++tot] = y;
Next[tot] = head[x];
head[x] = tot;
}
void dfs1(int x, int fa = -1) {
sz[x] = 1;
f[x] = fa;
int mx = 0;
for (int i = head[x]; i; i = Next[i]) {
int y = ver[i];
if(y == fa) continue;
d[y] = d[x] + 1;
dfs1(y, x);
sz[x] += sz[y];
if(sz[y] > mx) {
mx = sz[y];
son[x] = y;
}
}
}
void dfs2(int x, int fa, int t) {
dfn[x] = ++cnt;
top[x] = t;
if(son[x]) dfs2(son[x], x, t);
for (int i = head[x]; i; i = Next[i]) {
int y = ver[i];
if(y == fa || y == son[x]) continue;
dfs2(y, x, y);
}
}
void pushup(int o) {
tr[o].val = tr[ls(o)].val + tr[rs(o)].val;
}
void maintain(int o, int val) {
tr[o].val = val * (tr[o].r - tr[o].l + 1);
tr[o].lz = val;
}
void pushdown(int o) {
if(tr[o].lz != -1) {
maintain(ls(o), tr[o].lz);
maintain(rs(o), tr[o].lz);
tr[o].lz = -1;
}
}
void build(int o, int l, int r) {
tr[o].l = l, tr[o].r = r;
if(l == r) {
tr[o].val = 0;
tr[o].lz = -1;
return;
}
int mid = (l + r) >> 1;
build(ls(o), l, mid);
build(rs(o), mid + 1, r);
pushup(o);
}
void update(int o, int l, int r, int ql, int qr, int val) {
if(l >= ql && r <= qr) {
tr[o].val = (r - l + 1) * val;
tr[o].lz = val;
return;
}
pushdown(o);
int mid = (l + r) >> 1;
if(ql <= mid) update(ls(o), l, mid, ql, qr, val);
if(qr > mid) update(rs(o), mid + 1, r, ql, qr, val);
pushup(o);
}
int query(int o, int l, int r, int ql, int qr) {
if(l >= ql && r <= qr) {
return tr[o].val;
}
pushdown(o);
int mid = (l + r) >> 1, ans = 0;
if(ql <= mid) ans += query(ls(o), l, mid , ql, qr);
if(qr > mid) ans += query(rs(o), mid + 1, r, ql, qr);
return ans;
}
int solve(int x) {
int ans = 0, st = x;
while(x != -1) {
ans += query(1, 1, n, dfn[top[x]], dfn[x]);
x = f[top[x]];
}
return d[st] - d[0] + 1 - ans;
}
void update1(int x, int val) {
while(x != -1) {
update(1, 1, n, dfn[top[x]], dfn[x], val);
x = f[top[x]];
}
}
char s[110];
int main() {
int x, m;
scanf("%d", &n);
for (int i = 1; i < n; i++) {
scanf("%d", &x);
add(x, i);
add(i, x);
}
f[0] = -1;
build(1, 1, n);
dfs1(0);
dfs2(0, -1, 0);
scanf("%d", &m);
while(m--) {
scanf("%s", s + 1);
if(s[1] == ‘i‘) {
scanf("%d", &x);
int tmp = query(1, 1, n, dfn[x], dfn[x]);
if(tmp == 1) {
printf("0\n");
continue;
}
printf("%d\n", solve(x));
update1(x, 1);
} else {
scanf("%d", &x);
int tmp = query(1, 1, n, dfn[x], dfn[x]);
if(tmp == 0) {
printf("0\n");
continue;
}
printf("%d\n", query(1, 1, n, dfn[x], dfn[x] + sz[x] - 1));
update(1, 1, n, dfn[x], dfn[x] + sz[x] - 1, 0);
}
}
}
原文:https://www.cnblogs.com/pkgunboat/p/10645132.html