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BZOJ 4545

bzoj 4545

给定一个树,支持几种操作。

  • 本质不同子串询问
  • 加入子树
  • 询问字符串\(S\) 在树上的出现次数。

好码好码

重点就是维护\(parent\) 树,考虑用\(LCT\)维护此树。

第三问就是匹配点的\(right\)集合大小,算一算就可以了。

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 200010;

int read () {
	int q=0,f=1;char ch=getchar();
	while(!isdigit(ch)) {
		if(ch=='-')f=-1;ch=getchar();
	}
	while(isdigit(ch)){
		q=q*10+ch-'0';ch=getchar();
	}
	return q*f;
}
long long ans;
int opt,x,y,n,q;
struct LCT {
	int son[MAXN][2];
	int fa[MAXN];
	int v[MAXN];
	int tag[MAXN];
	int rev[MAXN];
	int getson(int x) {
		return son[fa[x]][1] == x;
	}
	int isroot(int x) {
		return son[fa[x]][0] != x and son[fa[x]][1] != x;
	}
	void down(int now) {
		swap(son[now][0],son[now][1]);
		rev[now] ^= 1;
	}
	void Add_val(int x,int k) {
		if(x) {
			v[x] += k;
			tag[x] += k;
		}
	}
	void pushdown(int now) {
		if(rev[now]) {
			down(son[now][0]);
			down(son[now][1]);
			rev[now] = 0;
		}
		if(tag[now]) {
			Add_val(son[now][0],tag[now]);
			Add_val(son[now][1],tag[now]);
			tag[now] = 0;
		}
	}
	void dfs(int now) {
		if(!isroot(now)) {
			dfs(fa[now]);
		}
		pushdown(now);
	}
	void rotate(int now) {
		int y = fa[now];
		int z = fa[y];
		int wht = getson(now);
		fa[now] = z;
		if(!isroot(y)) {
			son[z][y == son[z][1]] = now;
		}
		fa[son[now][wht ^ 1]] = y;
		son[y][wht] = son[now][wht ^ 1];
		fa[y] = now;
		son[now][wht ^ 1] = y;
	}
	void splay(int now) {
		dfs(now);
		for(int i = fa[now]; !isroot(now) ; rotate(now),i = fa[now]) {
			if(!isroot(i)) {
				rotate(getson(now) == getson(i) ? i : now);
			}
		}
	}
	void access(int now) {
		for(int i = 0;now;i = now,now = fa[now]) {
			splay(now);
			son[now][1] = i;
		}
	}
	void makeroot(int now) {
		access(now);
		splay(now);
		down(now);
	}
	void link(int x,int y) {
		makeroot(x);
		fa[x] = y;
	}
	void cut(int x,int y) {
		makeroot(x);
		access(y);
		splay(y);
		son[y][0] = fa[x] = 0;
	}
	void Add(int x,int y,int k) {
		makeroot(x);
		access(y);
		splay(y);
		Add_val(y,k);
	}
	int query(int now) {
		splay(now);
		return v[now];
	}
}lct;

struct SAM {
	int ch[MAXN][3];
	int len[MAXN];
	int fail[MAXN];
	int Cnt;
	SAM() {
		Cnt = 1;
	}
	int Copy(int c) {
		int now = ++Cnt;
		for(int i = 0;i <= 2; ++i) {
			ch[now][i] = ch[c][i];
		}
		return now;
	}
	void link(int x,int y) {
		fail[x] = y;
		lct.link(x,y);
		ans += len[x] - len[y];
	}
	int work(int p,int c) {
		int q = ch[p][c];
		int nq = Copy(q);
		len[nq] = len[p] + 1;
		lct.v[nq] = lct.query(q);
		lct.cut(q,fail[q]);
		ans -= len[q] - len[fail[q]];
		link(nq,fail[q]);
		link(q,nq);
		for( ; ch[p][c] == q ; p = fail[p]) {
			ch[p][c] = nq;
		}
		return nq;
	}
	int insert(int p,int c) {
		int cur;
		if(ch[p][c]) {
			cur = len[ch[p][c]] == len[p] + 1 ? ch[p][c] : work(p,c);
		}
		else {
			cur = ++Cnt;
			len[cur] = len[p] + 1;
			for( ; p and !ch[p][c] ; p = fail[p]) {
				ch[p][c] = cur;
			}
			if(!p) {
				link(cur,1);
			}
			else if(len[ch[p][c]] == len[p] + 1) {
				link(cur,ch[p][c]);
			}
			else {
				link(cur,work(p,c));
			}
		}
		lct.Add(cur,1,1);
		return cur;
	}
}sam;

struct graph {
	int head[MAXN];
	int cnt;
	graph () {
		cnt = 0;
	}
	int ver[MAXN];
	int nxt[MAXN];
	int val[MAXN];
	int vis[MAXN];
	int pos[MAXN];
	void add(int u,int v,int w) {
		ver[++cnt] = v;
		nxt[cnt] = head[u];
		head[u] = cnt;
		val[cnt] = w;
	}
	void Add(int u,int v,int w) {
		add(u,v,w);
		add(v,u,w);
	}
	void dfs(int now) {
		vis[now] = 1;
		for(int i = head[now];i;i=nxt[i]) {
			int y = ver[i];
			if(vis[y]) {
				continue;
			}
			pos[y] = sam.insert(pos[now],val[i]);
			dfs(y);
		}
	}
}G;
char s[MAXN];

int main () {
	x = read(),n = read();
	for(int i = 1;i < n; ++i) {
		x = read(),y = read();
		scanf("%s",s);
		G.Add(x,y,s[0] - 'a');
	}
	G.pos[1] = 1;
	G.dfs(1);
	q = read();
	while(q--) {
		opt = read();
		if(opt == 1) {
			cout<<ans<<endl;
		}
		else if(opt == 2) {
			x = read(),y = read();
			for(int i = 1;i < y; ++i) {
				int X = read(),Y = read();
				scanf("%s",s);
				G.Add(X,Y,s[0] - 'a');
			}
			G.dfs(x);
		}
		else {
			scanf("%s",s);int len = strlen(s);
			int p = 1;
			int tag = 1;
			for(int i = 0;i < len; ++i) {
				if(!sam.ch[p][s[i] - 'a']) {
					cout<<0<<endl;
					tag = 0;
					break;
				}
				p = sam.ch[p][s[i] - 'a'];
			}
			if(tag) {
				cout<<lct.query(p)<<endl;
			}
		}
	}
	return 0;
}
posted @ 2019-01-04 11:23  Allorkiya  阅读(213)  评论(0编辑  收藏  举报