(中等) UESTC 94 Bracket Sequence,线段树+括号。
There is a sequence of brackets, which supports two kinds of operations.
- we can choose a interval [l,r], and set all the elements range in this interval to left bracket or right bracket.
- we can reverse a interval, which means that for all the elements range in [l,r], if it's left bracket at that time, we change it into right bracket, vice versa.
Fish is fond of Regular Bracket Sequence
, so he want to know whether a interval [l,r] of the sequence is regular or not after doing some operations.
Let us define a regular brackets sequence in the following way:
- Empty sequence is a regular sequence.
- If
S
is a regular sequence, then(S)
is also a regular sequences. - If
A
andB
are regular sequences, thenAB
is a regular sequence.
题目大意就是说给你一个括号序列,对他进行操作和询问,包括反转和覆盖两个操作。
维护一个总和,还有一个最小前缀和(还要维护最大前缀和,在反转的时候计算最小的。)。当总和和最小前缀和都为0,则成立。
这个题又被坑了好久,没办法,水平太差了,错了十几次,电子科大的提交记录都被我刷屏了。。。错误百出。。。
代码如下:
#include<iostream>
#include<cstdio>
#include<cstring> #define lson L,M,po*2
#define rson M+1,R,po*2+1
#define max(a,b) (a>b?a:b)
#define min(a,b) (a<b?a:b) using namespace std; int BIT[*];
int QS[*];
int MS[*];
int XOR[*];
int COL[*];
char ss[]; void pushUP(int po)
{
BIT[po]=BIT[po*]+BIT[po*+];
QS[po]=max(QS[po*],BIT[po*]+QS[po*+]); //这里要注意。
MS[po]=min(MS[po*],BIT[po*]+MS[po*+]);
} void pushDown(int po,int len)
{
if(COL[po])
{
COL[po*]=COL[po];
COL[po*+]=COL[po];
XOR[po*]=XOR[po*+]=; //这里不能忘记。
BIT[po*]=(len-(len/))*COL[po];
BIT[po*+]=(len/)*COL[po]; QS[po*]=max(-,BIT[po*]);
QS[po*+]=max(-,BIT[po*+]);
MS[po*]=min(,BIT[po*]);
MS[po*+]=min(,BIT[po*+]); COL[po]=;
} if(XOR[po])
{
int temp; XOR[po*]=!XOR[po*];
XOR[po*+]=!XOR[po*+]; BIT[po*]=-BIT[po*];
BIT[po*+]=-BIT[po*+]; temp=QS[po*];
QS[po*]=-MS[po*];
MS[po*]=-temp; temp=QS[po*+];
QS[po*+]=-MS[po*+];
MS[po*+]=-temp; XOR[po]=;
}
} void build_tree(int L,int R,int po)
{
XOR[po]=;
COL[po]=; if(L==R)
{
if(ss[L]=='(')
{
BIT[po]=;
QS[po]=;
MS[po]=;
}
else
{
BIT[po]=-;
QS[po]=-;
MS[po]=-;
} return;
} int M=(L+R)/; build_tree(lson);
build_tree(rson); pushUP(po);
} void update_col(int ul,int ur,int ut,int L,int R,int po)
{
if(ul<=L&&ur>=R)
{
XOR[po]=;
COL[po]=ut;
BIT[po]=ut*(R-L+); QS[po]=max(-,BIT[po]);
MS[po]=min(,BIT[po]); return;
} pushDown(po,R-L+); int M=(L+R)/; if(ul<=M)
update_col(ul,ur,ut,lson);
if(ur>M)
update_col(ul,ur,ut,rson); pushUP(po);
} void update_xor(int ul,int ur,int L,int R,int po)
{
if(ul<=L&&ur>=R)
{
XOR[po]=!XOR[po];
BIT[po]=-BIT[po]; int temp=QS[po];
QS[po]=-MS[po];
MS[po]=-temp; return;
} pushDown(po,R-L+); int M=(L+R)/; if(ul<=M)
update_xor(ul,ur,lson);
if(ur>M)
update_xor(ul,ur,rson); pushUP(po);
} int query(int &qs,int ql,int qr,int L,int R,int po) //不能忘记写 & !!!
{
if(ql<=L&&qr>=R)
{
qs=MS[po];
return BIT[po];
} pushDown(po,R-L+); int M=(L+R)/;
int ans=; if(qr<=M)
return query(qs,ql,qr,lson);
if(ql>M)
return query(qs,ql,qr,rson); int temp1,temp2,a1; a1=query(temp1,ql,qr,lson);
ans=a1+query(temp2,ql,qr,rson); qs=min(temp1,temp2+a1); return ans;
} bool getans(int ql,int qr,int N)
{
int t1;
int ans; if((qr-ql)%==)
return ; ans=query(t1,ql,qr,,N,); if(ans==&&t1==)
return ;
else
return ;
} int main()
{
int T;
int N,Q;
char t1[],t2[];
int a,b;
cin>>T; for(int cas=;cas<=T;++cas)
{
printf("Case %d:\n",cas); scanf("%d",&N);
scanf("%s",ss); build_tree(,N-,); //这里应该是N-1。 scanf("%d",&Q); for(int i=;i<Q;++i)
{
scanf("%s %d %d",t1,&a,&b); if(t1[]=='s')
{
scanf("%s",t2);
update_col(a,b,t2[]=='('?:-,,N-,);
}
else if(t1[]=='r')
update_xor(a,b,,N-,);
else
if(getans(a,b,N-))
printf("YES\n");
else
printf("NO\n");
} printf("\n");
} return ;
}
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