洛谷题目链接：[NOI2015]软件包管理器
LOJ题目链接：[NOI2015]软件包管理器
题解原发于我的blog
首先，很明显这是一道树链剖分的题。
注意到一个软件只会以来一个软件，并且不会出现环，所以每次都可以连一条((x , i))的边。
当安装一个软件时，就把((1 , x))的路径上所有的点的转态变为(1)
但卸载一个软件时，就把(x)及它的所有的子树变为(0)
线段树维护即可
推荐一道树链剖分的好题【模板】树链剖分
做完这道题就可以差不多学完所有关于树链剖分的芝士
最后贴上代码知道你们只看这个：
#include <cstdio>
#include <algorithm>
using namespace std;
template <typename T>
inline void read(T &x)
{
x = 0;
char s = getchar();
bool f = false;
while (!(s >= '0' && s <= '9'))
{
if (s == '-')
f = true;
s = getchar();
}
while (s >= '0' && s <= '9')
{
x = (x << 1) + (x << 3) + s - '0';
s = getchar();
}
if (f)
x = (~x) + 1;
}
#define re register
#define ls (k << 1)
#define rs (k << 1 | 1)
const int N = 1e5 + 10, M = 2e5 + 10, T = 4e5 + 10;
struct Edge
{
int next, to;
} edge[M];
int num_edge, head[N];
struct Tree
{
int l, r, size, sum, flag;
} tree[T];
char s[110];
int n, q, cnt;
int idx[N], rk[N], dep[N], fa[N], top[N], size[N], son[N]; //rk[i]其实并没有什么用这只是我的习惯
inline void add_edge(int from, int to)
{
edge[++num_edge].next = head[from];
edge[num_edge].to = to;
head[from] = num_edge;
}
inline void dfs1(int u, int f)
{
fa[u] = f;
dep[u] = dep[f] + 1;
size[u] = 1;
for (re int i = head[u]; i; i = edge[i].next)
{
int &v = edge[i].to;
if (v == f)
continue;
dfs1(v, u);
size[u] += size[v];
if (size[son[u]] < size[v])
son[u] = v;
}
}
inline void dfs2(int u, int tp)
{
idx[u] = ++cnt;
rk[cnt] = u;
top[u] = tp;
if (!son[u])
return;
dfs2(son[u], tp);
for (re int i = head[u]; i; i = edge[i].next)
{
int &v = edge[i].to;
if (idx[v])
continue;
dfs2(v, v);
}
}
inline void pushdown(int k)
{
if (~tree[k].flag)//相当于tree[k].flag!=-1
{
tree[ls].sum = tree[ls].size * tree[k].flag;
tree[rs].sum = tree[rs].size * tree[k].flag;
tree[ls].flag = tree[rs].flag = tree[k].flag;
tree[k].flag = -1;
}
}
inline void build(int k, int l, int r)
{
tree[k].l = l;
tree[k].r = r;
tree[k].size = r - l + 1;
if (l == r)
{
tree[k].flag = -1;
return;
}
int mid = l + r >> 1;
build(ls, l, mid);
build(rs, mid + 1, r);
}
inline void update1(int k, int l, int r, int val)
{
if (l <= tree[k].l && tree[k].r <= r)
{
tree[k].sum = tree[k].size * val;
tree[k].flag = val;
return;
}
int mid = (tree[k].l + tree[k].r) >> 1;
pushdown(k);
if (l <= mid)
update1(ls, l, r, val);
if (mid < r)
update1(rs, l, r, val);
tree[k].sum = tree[ls].sum + tree[rs].sum;
}
inline void update2(int x, int y)
{
while (top[x] != top[y])
{
if (dep[top[x]] < dep[top[y]])
swap(x, y);
update1(1, idx[top[x]], idx[x], 1);
x = fa[top[x]];
}
if (dep[x] > dep[y])
swap(x, y);
update1(1, idx[x], idx[y], 1);
}
int main()
{
read(n);
for (re int i = 2, x; i <= n; ++i)
{
read(x);
add_edge(x + 1, i);
}
dfs1(1, 0);
dfs2(1, 1);
build(1, 1, n);
read(q);
for (re int i = 1, before, x; i <= q; ++i)
{
scanf("%s", s);
read(x);
++x;
before = tree[1].sum;
if (s[0] == 'i')
{
update2(1, x);
printf("%d
", tree[1].sum - before);
}
else if (s[0] == 'u')
{
update1(1, idx[x], idx[x] + size[x] - 1, 0);
printf("%d
", before - tree[1].sum);
}
}
return 0;
}
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