Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P5556 圣剑护符

P5556 圣剑护符

路径修改,树剖即可;路径询问是否存在两个子集异或和相同。两个子集异或和相同则一定存在一个集合异或和为 $0$ ,即不为一个线性基。线性基最多有 $30$ 个位置,那么考虑一条路径如果元素个数大于 $30$ ,那么答案一定为 $YES$ 。否则暴力插入线性基。复杂度 $O(n \log n \log a)$ 。

查看代码
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#include <cstdio>
#include <vector>
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write(Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar(x % 10 + '0');
}
const int N = 1e5 + 10, M = 30;
vector <int> g[N];
int n, m, w[N], d[N], p[N];
int stmp, dfn[N], rnk[N], sz[N], son[N], top[N];
struct Node
{
int l, r, v, tag;
} tr[N << 2];
void add (int x, int k)
{
tr[x].v ^= k, tr[x].tag ^= k;
}
void pushdown (int x)
{
add(x << 1, tr[x].tag), add(x << 1 | 1, tr[x].tag);
tr[x].tag = 0;
}
void build (int l = 1, int r = n, int x = 1)
{
tr[x].l = l, tr[x].r = r;
if (l == r)
return tr[x].v = w[rnk[l]], void();
int mid = l + r >> 1;
build(l, mid, x << 1) , build(mid + 1, r, x << 1 | 1);
}
void modify (int l, int r, int k, int x = 1)
{
if (tr[x].l > r || l > tr[x].r)
return;
if (tr[x].l >= l && tr[x].r <= r)
return add(x, k);
pushdown(x);
modify(l, r, k, x << 1), modify(l, r, k, x << 1 | 1);
}
int query (int t, int x = 1)
{
if (tr[x].l == tr[x].r)
return tr[x].v;
pushdown(x);
int mid = tr[x].l + tr[x].r >> 1;
return query(t, t <= mid ? x << 1 : x << 1 | 1);
}
void ModifyPath (int x, int y, int k)
{
while (top[x] ^ top[y])
{
d[top[x]] < d[top[y]] && (swap(x, y), 0);
modify(dfn[top[x]], dfn[x], k);
x = p[top[x]];
}
d[x] > d[y] && (swap(x, y), 0);
modify(dfn[x], dfn[y], k);
}
int lca (int x, int y)
{
while (top[x] ^ top[y])
{
d[top[x]] < d[top[y]] && (swap(x, y), 0);
x = p[top[x]];
}
d[x] > d[y] && (swap(x, y), 0);
return x;
}
void dfs1 (int x)
{
sz[x] = 1;
son[x] = -1;
for (int i : g[x])
if (!d[i])
{
d[i] = d[x] + 1;
p[i] = x;
dfs1(i);
sz[x] += sz[i];
if (son[x] == -1 || sz[i] > sz[son[x]])
son[x] = i;
}
}
void dfs2 (int x, int t)
{
rnk[dfn[x] = ++stmp] = x;
top[x] = t;
~son[x] && (dfs2(son[x], t), 0);
for (int i : g[x])
i ^ son[x] && i ^ p[x] && (dfs2(i, i), 0);
}
int bs[M];
bool insert (int x)
{
for (int i = M - 1; ~i; i--)
if (x >> i & 1)
{
if (!bs[i])
return bs[i] = x, true;
x ^= bs[i];
}
return false;
}
bool calc (int x, int y)
{
int t = lca(x, y);
if (d[x] + d[y] - 2 * d[t] + 1 > M)
return true;
for (int i = 0; i < M; i++)
bs[i] = 0;
if (!insert(query(dfn[t])))
return true;
while (x ^ t)
{
if (!insert(query(dfn[x])))
return true;
x = p[x];
}
while (y ^ t)
{
if (!insert(query(dfn[y])))
return true;
y = p[y];
}
return false;

}
int main ()
{
read(n), read(m);
for (int i = 1; i <= n; i++)
read(w[i]);
for (int i = 1, a, b; i < n; i++)
{
read(a), read(b);
g[a].push_back(b);
g[b].push_back(a);
}
d[1] = 1, dfs1(1), dfs2(1, 1);
build();
for (char op[8]; m; m--)
{
scanf("%s", op);
int x, y;
read(x), read(y);
if (op[0] == 'Q')
puts(calc(x, y) ? "YES" : "NO");
else if (op[0] == 'U')
{
int k;
read(k);
ModifyPath(x, y, k);
}
}
return 0;
}