【BZOJ 3735】苹果树 树上莫队(树分块+离线莫队+鬼畜的压行)

2016-05-09 UPD：学习了新的DFS序列分块，然后发现这个东西是战术核导弹？反正比下面的树分块不知道要快到哪里去了

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N = 50003;
const int M = 100003;
void read(int &k) {
	k = 0; int fh = 1; char c = getchar();
	for(; c < '0' || c > '9'; c = getchar())
		if (c == '-') fh = -1;
	for(; c >= '0' && c <= '9'; c = getchar())
		k = (k << 1) + (k << 3) + c - '0';
	k = k * fh;
}

struct node {int nxt, to;} E[N << 1];
struct quest {int l, r, id, a, b, lca;} Q[M];
int f[N][17], pos[N << 1], L[N], R[N], color[N], cal[N];
int point[N], bel[N << 1], ans = 0, cnt = 0, n, m, A[M], deep[N];
bool vis[N];

void ins(int x, int y) {E[++cnt].nxt = point[x]; E[cnt].to = y; point[x] = cnt;}
void _(int x, int fa) {
	pos[L[x] = ++cnt] = x;
	for(int i = 1; i <= 16; ++i) {f[x][i] = f[f[x][i - 1]][i - 1]; if (f[x][i] == 0) break;}
	for(int tmp = point[x]; tmp; tmp = E[tmp].nxt)
		if (E[tmp].to != fa) 
			{deep[E[tmp].to] = deep[x] + 1; f[E[tmp].to][0] = x; _(E[tmp].to, x);}
	pos[R[x] = ++cnt] = x;
}
int LCA(int x, int y) {
	if (deep[x] < deep[y]) swap(x, y);
	int d = deep[x] - deep[y];
	for(int i = 0; i <= 16; ++i) if (d & (1 << i)) x = f[x][i];
	if (x == y) return x;
	for(int i = 16; i >= 0; --i) if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
	return f[x][0];
}
bool cmp(quest A, quest B) {return bel[A.l] == bel[B.l] ? A.r < B.r : A.l < B.l;}

int check(int a, int b) {return cal[a] && cal[b] && a != b;}
void update(int x) {
	if (vis[x]) {if (!(--cal[color[x]])) --ans;}
	else {if (!(cal[color[x]]++)) ++ans;}
	vis[x] = !vis[x];
}

int main() {
	read(n); read(m);
	for(int i = 1; i <= n; ++i) read(color[i]);
	int lca, u, v, a, b;
	for(int i = 1; i <= n; ++i) {
		read(u); read(v);
		ins(u, v); ins(v, u);
	}
	
	cnt = 0;
	_(1, 0);
	for(int i = 1; i <= m; ++i) {
		read(u); read(v); read(Q[i].a); read(Q[i].b); Q[i].id = i;
		lca = LCA(u, v);
		if (L[u] > L[v]) swap(u, v);
		if (u != lca) {Q[i].l = R[u]; Q[i].r = L[v]; Q[i].lca = lca;}
		else {Q[i].l = L[u]; Q[i].r = L[v]; Q[i].lca = 0;}
	}
	
	int nn = n << 1, sq = sqrt(nn + 0.5), tmp = 0; cnt = 1;
	for(int i = 1; i <= nn; ++i) {
		bel[i] = tmp;
		++cnt; if (cnt > sq) cnt = 1, ++tmp;
	}
	
	sort(Q + 1, Q + m + 1, cmp);
	
	int l = 1, r = 0, tol, tor;
	for(int i = 1; i <= m; ++i) {
		tol = Q[i].l; tor = Q[i].r;
		while (l < tol) update(pos[l++]);
		while (l > tol) update(pos[--l]);
		while (r < tor) update(pos[++r]);
		while (r > tor) update(pos[r--]);
		if (Q[i].lca) update(Q[i].lca);
		A[Q[i].id] = ans - check(Q[i].a, Q[i].b);
		if (Q[i].lca) update(Q[i].lca);
	}
	
	for(int i = 1; i <= m; ++i)
		printf("%d
", A[i]);
	
	return 0;
}

学习了树上莫队，树分块后对讯问的$dfs序$排序，然后就可以滑动树链处理答案了。

关于树链的滑动，只需要特殊处理一下$LCA$就行了。

在这里一条树链保留下来给后面的链来转移的$now$的为这条树链上所有点除去$LCA$的颜色种数。因为如果要考虑$LCA$情况就太多了，不如单独考虑$LCA$。

转移后加上当前链的$LCA$进行统计，然后再去掉这个$LCA$更新一下$now$值给后面的链转移。

这都是我的理解，说的有点不清楚，具体请看vfk的题解 OTZ 虽然不是这道题，但是通过这篇博客学习树上莫队也是很好的。

PS：压行大法使代码看起来像一堵墙

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 50003
#define M 100003
#define read(x) x=getint()
using namespace std;
inline int getint() {int k = 0, fh = 1; char c = getchar();	for(; c < '0' || c > '9'; c = getchar()) if (c == '-') fh = -1; for(; c >= '0' && c <= '9'; c = getchar()) k = k * 10 + c - '0'; return k * fh;}
int n, m, color[N], cnt = 0, fa[N][16], deep[N], dfn[N << 1], now = 0;
int belong[N], cntblo = 0, sqrblo, top = 0, sta[N], ans[M], colsum[N], point[N];
short v[N];
struct Enode {int nxt, to;} E[N << 1];
struct node {int x, y, a, b, id;} q[M];
inline bool cmp(node A, node B) {return belong[A.x] == belong[B.x] ? dfn[A.y] < dfn[B.y] : dfn[A.x] < dfn[B.x];}
inline void ins(int x, int y) {E[++cnt].nxt = point[x]; E[cnt].to = y; point[x] = cnt;}

inline void dfs(int x) {
	dfn[x] = ++cnt;
	int mark = top;
	for(int i = 1; i <= 15; ++i)
		fa[x][i] = fa[fa[x][i - 1]][i - 1];
	for(int tmp = point[x]; tmp; tmp = E[tmp].nxt) {
		int v = E[tmp].to;
		if (v == fa[x][0]) continue;
		deep[v] =deep[x] + 1;
		fa[v][0] = x;
		dfs(v);
		if (top - mark >= sqrblo) {
			++cntblo;
			while (top != mark)
				belong[sta[top--]] = cntblo;
		}
	}
	sta[++top] = x; 
}

inline int LCA(int x, int y) {
	if (deep[x] < deep[y])
		swap(x, y);
	int k = deep[x] - deep[y];
	for(int j = 15; j >= 0; --j)
		if (k & (1 << j))
			x = fa[x][j];
	if (x == y) return x;
	for(int j = 15; j >= 0; --j)
		if (fa[x][j] != fa[y][j])
			x = fa[x][j], y = fa[y][j];
	return fa[x][0];
}

inline void pushup(int x) {
	if (v[x]) {
		--colsum[color[x]];
		if (!colsum[color[x]])
			--now;
	} else {
		if (!colsum[color[x]])
			++now;
		++colsum[color[x]];
	}
	v[x] ^= 1;
}

inline void change(int x, int y) {
	while (x != y) {
		if (deep[x] < deep[y])
			pushup(y), y = fa[y][0];
		else
			pushup(x), x = fa[x][0];
	} //O)Z这个方法好神啊！！！我为什么想不到一个一个往上跳呢QAQ 
}

int main() {
	read(n); read(m);
	for(int i = 1; i <= n; ++i)
		read(color[i]);
	int u, v;
	for(int i = 1; i <= n; ++i) {
		read(u); read(v);
		if (u == 0 || v == 0) continue;
		ins(u, v);
		ins(v, u);
	}
	sqrblo = ceil(sqrt(n));
	cnt = 0;
	dfs(1);
	while (top)
		belong[sta[top--]] = cntblo;
	
	for(int i = 1; i <= m; ++i) {
		read(q[i].x); read(q[i].y); read(q[i].a); read(q[i].b); q[i].id = i;
		if (dfn[q[i].x] > dfn[q[i].y])
			swap(q[i].x, q[i].y);
	}
	
	sort(q + 1, q + m + 1, cmp);
	q[0].x = q[0].y = 1;
	
	for(int i = 1; i <= m; ++i) {
		change(q[i - 1].x, q[i].x);
		change(q[i - 1].y, q[i].y);
		int lca = LCA(q[i].x, q[i].y);
		pushup(lca);
		ans[q[i].id] = now;
		if (colsum[q[i].a] && colsum[q[i].b] && q[i].a != q[i].b)
			--ans[q[i].id];
		pushup(lca);
	}
	
	for(int i = 1; i <= m; ++i)
		printf("%d
", ans[i]);
	return 0;
}

$SDOI2016 Round1$之前做的最后一道题了，希望省选不要爆零啊$QAQ$

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原文地址：https://www.cnblogs.com/abclzr/p/5365666.html