牛客挑战赛46 F
可以很显然的发现，答案等于\(\sum{dist_{i,root}}-all(lca(l,l+1,l+2,,,,r))\)其中
我们可以使用线段树取\([l,r]lca\)。
每一个点到根的和，我们考虑先离线，再\(access\)过程。每一次打通一条链，就要把这条链上，已经遍历过的点到该链的链头的距离给剪掉。在随便使用一个数据结构维护前缀和单点修改即可。因为离线，存储R,每一次access后，一定是向后满足条件的，所以每一次相当于询问前缀是否可行。
#include<iostream>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<cstdlib>
#include<climits>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<bitset>
#include<map>
//#include<regex>
#include<cstdio>
#include <iomanip>
#include<unordered_map>
#pragma GCC diagnostic error "-std=c++11"
#pragma GCC optimize("-fdelete-null-pointer-checks,inline-functions-called-once,-funsafe-loop-optimizations,-fexpensive-optimizations,-foptimize-sibling-calls,-ftree-switch-conversion,-finline-small-functions,inline-small-functions,-frerun-cse-after-loop,-fhoist-adjacent-loads,-findirect-inlining,-freorder-functions,no-stack-protector,-fpartial-inlining,-fsched-interblock,-fcse-follow-jumps,-fcse-skip-blocks,-falign-functions,-fstrict-overflow,-fstrict-aliasing,-fschedule-insns2,-ftree-tail-merge,inline-functions,-fschedule-insns,-freorder-blocks,-fwhole-program,-funroll-loops,-fthread-jumps,-fcrossjumping,-fcaller-saves,-fdevirtualize,-falign-labels,-falign-loops,-falign-jumps,unroll-loops,-fsched-spec,-ffast-math,Ofast,inline,-fgcse,-fgcse-lm,-fipa-sra,-ftree-pre,-ftree-vrp,-fpeephole2",3)
#pragma GCC target("avx","sse2")
#define up(i,a,b) for(int i=a;i<b;i++)
#define dw(i,a,b) for(int i=a;i>b;i--)
#define upd(i,a,b) for(int i=a;i<=b;i++)
#define dwd(i,a,b) for(int i=a;i>=b;i--)
//#define local
typedef long long ll;
typedef unsigned long long ull;
const double esp = 1e-6;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int inf = 1e9;
using namespace std;
ll read()
{
char ch = getchar(); ll x = 0, f = 1;
while (ch<'0' || ch>'9') { if (ch == '-')f = -1; ch = getchar(); }
while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); }
return x * f;
}
typedef pair<int, int> pir;
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lrt root<<1
#define rrt root<<1|1
const int N = 2e5 + 10;
int n, Q;
vector<int>vec[N];
vector<pir>ques[N];
int dep[N * 2], dfn[N * 2], rk[N * 2], st, dp[N * 2][30];
int ans[N];
struct sgt {
int sum[N << 2];
void pushup(int root) {
sum[root] = sum[lrt] + sum[rrt];
}
void update(int l, int r, int root, int pos, int val) {
if (l == r) {
sum[root] += val; return;
}
int mid = (l + r) >> 1;
if (pos <= mid)update(lson, pos, val);
else update(rson, pos, val);
pushup(root);
}
int query(int l, int r, int root, int lf, int rt) {
if (lf > rt)return 0;
if (lf <= l && r <= rt) { return sum[root]; }
int mid = (l + r) >> 1;
int ans = 0;
if (lf <= mid)ans += query(lson, lf, rt);
if (rt > mid)ans += query(rson, lf, rt);
return ans;
}
}T;
struct LCT {
int ch[N][2], par[N], lz[N], val[N], sz[N];
int ident(int rt, int fa) {
return ch[fa][1] == rt;
}
void connect(int rt, int fa, int son) {
par[rt] = fa;
ch[fa][son] = rt;
}
bool isroot(int rt) {
int f = par[rt];
return ch[f][1] != rt && ch[f][0] != rt;
}
void rotate(int rt) {
int f = par[rt]; int ff = par[f]; int k = ident(rt, f);
connect(ch[rt][k ^ 1], f, k);
par[rt] = ff;
if (!isroot(f))ch[ff][ident(f, ff)] = rt;
connect(f, rt, k ^ 1);
pushup(f); pushup(rt);
}
void pushup(int rt) {
sz[rt] = sz[ch[rt][0]] + sz[ch[rt][1]] + 1;
}
void pushdown(int rt) {
if (lz[rt]) {
int ls = ch[rt][0];
int rs = ch[rt][1];
if (ls)val[ls] = lz[ls] = lz[rt];
if (rs)val[rs] = lz[rs] = lz[rt];
lz[rt] = 0;
}
}
void pushall(int rt) {
if (!isroot(rt))pushall(par[rt]);
pushdown(rt);
}
void splay(int x) {
pushall(x);
while (!isroot(x)) {
int f = par[x]; int ff = par[f];
if (!isroot(f))ident(f, ff) ^ ident(x, f) ? rotate(x) : rotate(f);
rotate(x);
}
}
void access(int x, int col) {
int y;
for (y = 0; x; y = x, x = par[x]) {
splay(x);
if (val[x])
T.update(1, n, 1, val[x], -sz[ch[x][0]] - 1);
T.update(1, n, 1, col, sz[ch[x][0]] + 1);
//cout <<val[x]<<" "<< T.query(1, n, 1, 1, val[x]) << " ";
ch[x][1] = y;
pushup(x);
}
val[y] = lz[y] = col;
}
}lct;
void dfs1(int u, int fa, int d) {
lct.par[u] = fa;
dfn[++st] = u; dep[st] = d;
rk[u] = st;
for (auto k : vec[u]) {
if (k == fa)continue;
dfs1(k, u, d + 1);
dfn[++st] = u;
dep[st] = d;
rk[u] = st;
}
}
void initst() {
upd(i, 0, 2 * n)dp[i][0] = i;
upd(j, 1, 30) {
for (int i = 1; i + (1 << j) - 1 <= 2 * n; i++) {
int t1 = dp[i][j - 1]; int t2 = dp[i + (1 << (j - 1))][j - 1];
dp[i][j] = dep[t1] < dep[t2] ? t1 : t2;
}
}
}
int lca(int x, int y) {
x = rk[x], y = rk[y];
if (x > y)swap(x, y);
int k = int(log2(y - x + 1));
int t1 = dp[x][k]; int t2 = dp[y - (1 << k) + 1][k];
return dep[t1] < dep[t2] ? dfn[t1] : dfn[t2];
}
struct LCAT {
int LCA[N << 2];
void pushup(int root) {
LCA[root] = lca(LCA[lrt], LCA[rrt]);
}
void build(int l, int r, int root) {
if (l == r) {
LCA[root] = l; return;
}
int mid = (l + r) >> 1;
build(lson); build(rson); pushup(root);
}
int query(int l, int r, int root, int lf, int rt) {
if (lf <= l && r <= rt) {
return LCA[root];
}
int mid = (l + r) >> 1;
int temp = -1;
if (lf <= mid)temp = query(lson, lf, rt);
if (rt > mid) {
if (temp == -1)temp = query(rson, lf, rt);
else temp = lca(temp, query(rson, lf, rt));
}
return temp;
}
}LCAT;
int main() {
n = read(); Q = read();
int u, v;
upd(i, 1, n - 1) {
u = read(), v = read();
vec[u].push_back(v);
vec[v].push_back(u);
}
upd(i, 1, Q) {
u = read(), v = read();
ques[v].push_back({ u,i });
}
dfs1(1, 0, 0);
initst();
upd(i, 1, n) {
lct.sz[i] = 1;
}
LCAT.build(1, n, 1);
//cout << LCAT.query(1, n, 1, 1, 2);
upd(i, 1, n) {
lct.access(i, i);
////cout << T.query(1, n, 1, 1, i);
//upd(j, 1, i) {
// cout << T.query(1, n, 1, 1, j) << " ";
//}cout << endl;
for (auto k : ques[i]) {
ans[k.second] = T.query(1, n, 1, k.first, i) - dep[rk[LCAT.query(1, n, 1, k.first, i)]] - 1;
}
}
upd(i, 1, Q)printf("%d\n", ans[i]);
return 0;
}
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