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【BZOJ】2329: [HNOI2011]括号修复(splay+特殊的技巧)

http://www.lydsy.com/JudgeOnline/problem.php?id=2329

和前一题一样,不就多了个replace操作吗。好,就打一下。

然后交上去wa了。。。。。。。。。。。。。。。。。。。。

看了题解,好神奇!

记住:以后pushdown的tag要考虑先后顺序!

因为invert和swap操作谁先谁后没有关系,那么考虑invert和replace这两个有冲突的关系

为什么有冲突呢?因为假如你replace的标记在先,invert标记在后,但是invert在pushdown里先执行QAQ,比如序列(()),本来replace了[2, 3] (,变成 (((),然后invert[2, 3], 应该变成 ()))。

可是!你pushdown里执行的是invert先,即本来(()),然后变成了()(),然后再replace,变成((()啦!!!所以错了!!!

解决方法就是,在执行invert后,改变replace的tag。

即在变成()()后,replace[2, 3] ( 变成 replace[2, 3] ),那么就是()))。!!!

#include <cstdio>#include <cstring>#include <cmath>#include <string>#include <iostream>#include <algorithm>#include <queue>#include <set>#include <map>using namespace std;typedef long long ll;#define rep(i, n) for(int i=0; i<(n); ++i)#define for1(i,a,n) for(int i=(a);i<=(n);++i)#define for2(i,a,n) for(int i=(a);i<(n);++i)#define for3(i,a,n) for(int i=(a);i>=(n);--i)#define for4(i,a,n) for(int i=(a);i>(n);--i)#define CC(i,a) memset(i,a,sizeof(i))#define read(a) a=getint()#define print(a) printf("%d", a)#define dbg(x) cout << (#x) << " = " << (x) << endl#define error(x) (!(x)?puts("error"):0)#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }const int oo=~0u>>1, N=1e5+10;struct node *null;struct node {	node *f, *c[2];	int s, k, lmin, lmax, rmin, rmax, sum, fix;	bool tag, rev;	node(int _k=0) { s=1; lmin=lmax=rmin=rmax=k=sum=_k; f=c[0]=c[1]=null; tag=rev=0; fix=0; }	void setc(node *x, bool d) { c[d]=x; x->f=this; }	bool d() { return f->c[1]==this; }	void pushup() {		s=c[0]->s+c[1]->s+1;		sum=c[0]->sum+c[1]->sum+k;		lmin=c[0]->lmin;		lmax=c[0]->lmax;		rmin=c[1]->rmin;		rmax=c[1]->rmax;		int mnl=min(0, c[1]->lmin), mnr=min(0, c[0]->rmin);		int mxl=max(0, c[1]->lmax), mxr=max(0, c[0]->rmax);		lmin=min(lmin, c[0]->sum+k+mnl);		lmax=max(lmax, c[0]->sum+k+mxl);		rmin=min(rmin, c[1]->sum+k+mnr);		rmax=max(rmax, c[1]->sum+k+mxr);	}	void upd1() {		if(this==null) return;		if(fix) fix=-fix;		tag=!tag; int t;		t=lmin; lmin=-lmax; lmax=-t;		t=rmin; rmin=-rmax; rmax=-t;		k=-k;		sum=-sum;	}	void upd2() {		if(this==null) return;		rev=!rev;		swap(c[0], c[1]);		swap(lmin, rmin);		swap(lmax, rmax);	}	void upd3(int _k) {		if(this==null) return;		fix=_k;		k=_k;		sum=s*k;		lmin=rmin=k<0?sum:k;		lmax=rmax=k<0?k:sum;	}	void pushdown() {		if(tag) {			c[0]->upd1();			c[1]->upd1();			tag=0;		}		if(rev) {			c[0]->upd2();			c[1]->upd2();			rev=0;		}		if(fix) {			c[0]->upd3(fix);			c[1]->upd3(fix);			fix=0;		}	}}*root;void PN(node *x) {	printf("key:%c \t lmin:%d \t lmax:%d \t rmin:%d \t rmax:%d \t sum:%d \t \n", x->k==1?‘(‘:(x->k==-1?‘)‘:‘N‘), x->lmin, x->lmax, x->rmin, x->rmax, x->sum);}void Pr(node *x) {	if(x==null) return;	x->pushdown();	Pr(x->c[0]);	if(x->k) printf("%c", x->k==1?‘(‘:‘)‘);	Pr(x->c[1]);}void P(node *x=root) {	Pr(x); puts("");}void rot(node *x) {	node *f=x->f;	f->pushdown(); x->pushdown(); bool d=x->d();	f->f->setc(x, f->d());	f->setc(x->c[!d], d);	x->setc(f, !d);	f->pushup();	if(f==root) root=x;}void splay(node *x, node *f=null) {	x->pushdown();	while(x->f!=f)		if(x->f->f==f) rot(x);		else x->d()==x->f->d()?(rot(x->f), rot(x)):(rot(x), rot(x));	x->pushup();}node *sel(node *x, int k) {	if(x==null) return x;	x->pushdown();	int s=x->c[0]->s;	if(s==k) return x;	if(s<k) return sel(x->c[1], k-s-1);	return sel(x->c[0], k);}node *getrange(int l, int r) {	splay(sel(root, l-1));	splay(sel(root, r+1), root);	return root->c[1]->c[0];}void fix1() {	int l=getint(), r=getint();	node *x=getrange(l, r);	x->pushdown();	x->upd1();}void fix2() {	int l=getint(), r=getint();	node *x=getrange(l, r);	x->pushdown();	x->upd2();}void fix3() {	int l=getint(), r=getint();	node *x=getrange(l, r);	char c=getchar(); while(c!=‘)‘&&c!=‘(‘) c=getchar();	int fix=c==‘)‘?-1:1;	x->pushdown();	x->upd3(fix);}void ask() {	int l=getint(), r=getint();	node *x=getrange(l, r);	l=(-x->lmin+1)>>1;	r=(x->rmax+1)>>1;	printf("%d\n", l+r);}char s[N];int n, m, q;void build(int l, int r, node *&x) {	if(l>r) return;	int mid=(l+r)>>1;	x=new node(s[mid]==‘)‘?-1:1);	if(l==r) return;	build(l, mid-1, x->c[0]);	build(mid+1, r, x->c[1]);	if(l<=mid-1) x->c[0]->f=x;	if(mid+1<=r) x->c[1]->f=x;	x->pushup();}void in(node *x) {	x->lmax=x->rmax=-oo;	x->lmin=x->rmin=oo;	x->sum=0;	x->k=0;}void init() {	null=new node(); null->s=0; in(null);	null->f=null->c[0]=null->c[1]=null;	root=new node(); in(root);	root->setc(new node(), 1); in(root->c[1]);	node *x;	read(n); read(m);	scanf("%s", s+1);	build(1, n, x);	root->c[1]->setc(x, 0);	root->c[1]->pushup();	root->pushup();}char o[10];int main() {	init();	while(m--) {		scanf("%s", o);		if(o[0]==‘Q‘) ask();		else if(o[0]==‘I‘) fix1();		else if(o[0]==‘S‘) fix2();		else fix3();		//P();	}	return 0;}

  

 


 

 

Description

Input

 

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Sample Input

 

Sample Output

 

HINT

 

Source

Splay

 

【BZOJ】2329: [HNOI2011]括号修复(splay+特殊的技巧)