| Time Limit: 3000MS | Memory Limit: 65536KB | 64bit IO Format: %lld & %llu |
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
维护每个结点的sum max，为方便reserv再多维护一个min reserv时交换max，min的绝对值再成-1
set与reserv并存时，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: ",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 }