[LNOI2014] LCA
题面
题解
水题。
对于一个区间\([l,r]\)的答案。
我们把它差分成\([1,r] - [1,l - 1]\)
然后因为右端点是固定的，用树剖+只有1个端点的莫队就解决了...
你谷恶评实在太严重了。
#include <bits/stdc++.h>
const int maxn = 5e4 + 10;
const int mod = 201314;
template<class t> inline void read(t& res) {
res = 0; char ch = getchar(); bool neg = 0;
while (!isdigit(ch))
neg |= ch == '-', ch = getchar();
while (isdigit(ch))
res = (res << 1) + (res << 3) + (ch & 15), ch = getchar();
if (neg)
res = -res;
}
int q, n, m, i, j, k, cntq;
int hd[maxn], ver[maxn << 1], nxt[maxn << 1], ans[maxn];
struct Questions {
int p, z, id; int type;
Questions() { p = z = type = id = 0; }
Questions(int _p, int _z, int i, int _t) { p = _p; z = _z; id = i; type = _t; }
inline friend bool operator < (Questions a, Questions b) { return a.p < b.p; }
} Q[maxn << 1];
inline void adde(int u, int v) {
static int cnte = 0;
ver[++cnte] = v; nxt[cnte] = hd[u];
hd[u] = cnte; return;
}
int s[maxn << 2], tag[maxn << 2];
inline void push_up(int u) { s[u] = (s[u << 1] + s[u << 1 | 1]) % mod; }
inline void down(int u, int l, int r, int v) { s[u] += (r - l + 1) * v; tag[u] += v; }
inline void push_down(int u, int l, int r) {
int mid = (l + r) >> 1;
down(u << 1, l, mid, tag[u]);
down(u << 1 | 1, mid + 1, r, tag[u]);
tag[u] = 0;
}
void segt_modify(int ml, int mr, int l, int r, int u, int v) {
// printf("%d %d %d %d %d %d\n", ml, mr, l, r, u, v);
if (ml <= l && r <= mr)
return s[u] += (r - l + 1) * v, tag[u] += v, void();
int mid = (l + r) >> 1;
push_down(u, l, r);
if (ml <= mid)
segt_modify(ml, mr, l, mid, u << 1, v);
if (mid < mr)
segt_modify(ml, mr, mid + 1, r, u << 1 | 1, v);
push_up(u);
}
int segt_query(int ql, int qr, int l, int r, int u) {
if (ql <= l && r <= qr)
return s[u];
int mid = (l + r) >> 1, res = 0;
push_down(u, l, r);
if (ql <= mid)
res += segt_query(ql, qr, l, mid, u << 1) % mod;
res %= mod;
if (mid < qr)
res += segt_query(ql, qr, mid + 1, r, u << 1 | 1) % mod;
res %= mod;
return res;
}
namespace Tree_Chain {
int hsn[maxn], sz[maxn], dep[maxn], fa[maxn], top[maxn];
int dfn[maxn], tim;
void dfs1(int u, int pa) {
dep[u] = dep[pa] + 1;
fa[u] = pa;
sz[u] = 1;
for (int i = hd[u]; ~i; i = nxt[i]) {
int v = ver[i];
if (v == pa)
continue;
dfs1(v, u);
sz[u] += sz[v];
if (sz[v] > sz[ hsn[u] ])
hsn[u] = v;
}
}
void dfs2(int u, int up) {
top[u] = up; dfn[u] = ++tim;
if (hsn[u])
dfs2(hsn[u], up);
for (int i = hd[u]; ~i; i = nxt[i]) {
int v = ver[i];
if (v == fa[u] || v == hsn[u])
continue;
dfs2(v, v);
}
}
void modify(int u, int v, int val) {
while (top[u] != top[v]) {
if(dep[ top[u] ] < dep[ top[v] ])
std::swap(u, v);
segt_modify(dfn[ top[u] ], dfn[u], 1, n, 1, val);
u = fa[ top[u] ];
}
if (dep[u] > dep[v])
std::swap(u, v);
segt_modify(dfn[u], dfn[v], 1, n, 1, val);
}
int query(int u, int v) {
int res = 0;
while (top[u] != top[v]) {
//printf("%d %d\n", u, v);
if (dep[ top[u] ] < dep[ top[v] ])
std::swap(u, v);
res += segt_query(dfn[ top[u] ], dfn[u], 1, n, 1) % mod;
res %= mod;
u = fa[ top[u] ];
}
if (dep[u] > dep[v])
std::swap(u, v);
res += segt_query(dfn[u], dfn[v], 1, n, 1) % mod;
res %= mod;
return res;
}
}
int main() {
memset(hd, -1, sizeof(hd));
read(n); read(q);
for (int i = 2, p; i <= n; i++)
read(p), p++, adde(p, i), adde(i, p);
Tree_Chain::dfs1(1, 0);
Tree_Chain::dfs2(1, 1);
for (int i = 1, l, r, z; i <= q; i++) {
read(l); read(r); read(z); z++;
Q[++cntq] = Questions(l, z, i, -1);
Q[++cntq] = Questions(r + 1, z, i, 1);
}
std::sort(Q + 1, Q + cntq + 1);
int p = 1;
for (int i = 1; i <= cntq; i++) {
while (p <= Q[i].p)
Tree_Chain::modify(1, p++, 1);
ans[ Q[i].id ] += Tree_Chain::query(1, Q[i].z) * Q[i].type;
}
for (int i = 1; i <= q; i++)
printf("%d\n", (ans[i] + mod) % mod);
return 0;
}