UESTC 1546 Bracket Sequence

                                    Bracket Sequence

Time Limit: 3000MS

Memory Limit: 65536KB

64bit IO Format: %lld & %llu

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Description

There is a sequence of brackets, which supports two kinds of operations.
1. we can choose a interval [l,r], and set all the elements range in this interval to left bracket or right bracket. 
2. 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 opearations.

Let us define a regular brackets sequence in the following way: 

1. Empty sequence is a regular sequence. 
2. If S is a regular sequence, then (S) is also a regular sequences. 
3. If A and B are regular sequences, then AB is a regular sequence.

Input

In the first line there is an integer T (T≤10), indicates the number of test cases. Each case begins with a line containing an integers N (N ≤ 100,000 and N is a even number), the size of the initial brackets sequence. The next line contains a string whose length is N consisting of '(' and ')'. In the third of each test case, there is an integer M(M ≤ 100,000) indicates the number of queries. Each of the following M lines contains one operation as mentioned below. The index of the bracket sequence is labeled from 0 to N - 1.

Three operation description:
set l r c: change all the elements range in [l,r] into '(' or ')'.(c is '(' or ')')
reverse l r: reverse the interval [l,r]
query l,r: you should answer that interval [l,r] is regular or not

Output

For each test case, print a line containing the test case number (beginning with 1) on its own line, then the answer for each "query" operation, if the interval is regular, print "YES", otherwise print "NO", one on each line.
Print a blank line after each test case.

Sample Input

1
6
((()))
8
query 0 5
set 0 5 (
query 0 5
reverse 3 5
query 0 5
query 1 4
query 2 3
query 0 4

Sample Output

Case 1:
YES
NO
YES
YES
YES
NO

Hint

Huge input, use "scanf" instead of "cin".

Source

Classic Problem

 

線段樹，把左右括號標記成-1,1。。。合法的區間的總和爲零，且從左向右的累加和 小於等於 0
維護每一個結點的sum max，爲方便reserv再多維護一個min reserv時交換max，min的絕對值再成-1
set與reserv並存時，要先reserv再set

 

  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdio>
  4 
  5 #define lson l,m,rt<<1
  6 #define rson m+1,r,rt<<1|1
  7 
  8 using namespace std;
  9 
 10 const int maxn=111111;
 11 
 12 int setv[maxn<<2],rev[maxn<<2],sum[maxn<<2],mx[maxn<<2],mi[maxn<<2];
 13 char str[maxn];
 14 
 15 void reserve(int rt)
 16 {
 17     sum[rt]=-sum[rt];
 18     swap(mx[rt],mi[rt]);
 19     mx[rt]=-mx[rt];
 20     mi[rt]=-mi[rt];
 21     rev[rt]^=1;
 22 }
 23 
 24 void set(int op,int rt,int len)
 25 {
 26     sum[rt]=op*len;
 27     mx[rt]=sum[rt]>0?sum[rt]:0;
 28     mi[rt]=sum[rt]<0?sum[rt]:0;
 29     setv[rt]=op;rev[rt]=0;
 30 }
 31 
 32 void push_down(int rt,int ll,int lr)
 33 {
 34     if(rev[rt])
 35     {
 36         if(setv[rt]) setv[rt]*=-1;
 37         else reserve(rt<<1),reserve(rt<<1|1);
 38         rev[rt]=0;
 39     }
 40     if(setv[rt])
 41     {
 42         set(setv[rt],rt<<1,ll),set(setv[rt],rt<<1|1,lr);
 43         setv[rt]=0;
 44     }
 45 }
 46 
 47 void push_up(int rt)
 48 {
 49     sum[rt]=sum[rt<<1]+sum[rt<<1|1];
 50     mx[rt]=max(mx[rt<<1],sum[rt<<1]+mx[rt<<1|1]);
 51     mi[rt]=min(mi[rt<<1],sum[rt<<1]+mi[rt<<1|1]);
 52 }
 53 
 54 void build(int l,int r,int rt)
 55 {
 56     setv[rt]=rev[rt]=0;
 57     if(l==r)
 58     {
 59         sum[rt]=(str[l]=='(')?-1:1;
 60         mx[rt]=(sum[rt]<0)?0:1;
 61         mi[rt]=(sum[rt]<0)?-1:0;
 62         return;
 63     }
 64     int m=(l+r)>>1;
 65     build(lson),build(rson);
 66     push_up(rt);
 67 }
 68 
 69 void update(int L,int R,int c,int l,int r,int rt)
 70 {
 71     if(L<=l&&r<=R)
 72     {
 73         if(c) set(c,rt,r-l+1);
 74         else reserve(rt);
 75         return ;
 76     }
 77     int m=(l+r)>>1;
 78     push_down(rt,m-l+1,r-m);
 79     if(L<=m) update(L,R,c,lson);
 80     if(R>m) update(L,R,c,rson);
 81     push_up(rt);
 82 }
 83 
 84 int query_sum(int L,int R,int l,int r,int rt)
 85 {
 86     if(L<=l&&r<=R)
 87     {
 88         return sum[rt];
 89     }
 90     int m=(l+r)>>1,ans=0;
 91     push_down(rt,m-l+1,r-m);
 92     if(L<=m) ans+=query_sum(L,R,lson);
 93     if(R>m) ans+=query_sum(L,R,rson);
 94     push_up(rt);
 95     return ans;
 96 }
 97 
 98 int query_max(int L,int R,int l,int r,int rt)
 99 {
100     if(L<=l&&r<=R)
101     {
102         return mx[rt];
103     }
104     int m=(l+r)>>1,ret;
105     push_down(rt,m-l+1,r-m);
106     if(R<=m) ret=query_max(L,R,lson);
107     else if(L>m) ret=query_max(L,R,rson);
108     else ret=max(query_max(L,R,lson),query_sum(L,R,lson)+query_max(L,R,rson));
109     push_up(rt);
110     return ret;
111 }
112 
113 int main()
114 {
115     int t,cas=1;
116     char cmd[20];
117     scanf("%d",&t);
118     while(t--)
119     {
120         int n,m;
121         printf("Case %d:\n",cas++);
122         scanf("%d",&n);
123         scanf("%s",str);
124         build(0,n-1,1);
125         scanf("%d",&m);
126         while(m--)
127         {
128             scanf("%s",cmd);
129             if(cmd[0]=='s')
130             {
131                 int a,b; char c[10];
132                 scanf("%d%d%s",&a,&b,c);
133                 if(c[0]=='(')
134                     update(a,b,-1,0,n-1,1);
135                 else if(c[0]==')')
136                     update(a,b,1,0,n-1,1);
137             }
138             else if(cmd[0]=='r')
139             {
140                 int a,b;
141                 scanf("%d%d",&a,&b);
142                 update(a,b,0,0,n-1,1);
143             }
144             else if(cmd[0]=='q')
145             {
146                 int a,b;
147                 scanf("%d%d",&a,&b);
148                 if(!query_sum(a,b,0,n-1,1)&&!query_max(a,b,0,n-1,1)) puts("YES");
149                 else puts("NO");
150             }
151         }
152         putchar(10);
153     }
154     return 0;
155 }