题意：给出一个字符串，只有 ( ) 这两种，有三种操作
1、query查找一段字符里的括号是否匹配
2、set一段字符为某一种
3、reverse 把一段中的括号都变为其反
解法：用1表示该位置的'(' ,-1表示')'
线段树中的重要变量sum-和，lmin该段所有前缀中最小和，lmax前缀中最大和，br记录reverse的次数
t[cur].sum=t[lson].sum+t[rson].sum;
t[cur].lmin=min(t[lson].lmin,t[rson].lmin+t[lson].sum);
t[cur].lmax=max(t[lson].lmax,t[son].lmax+t[lson].sum);
注意：向下更新的顺序，。。应该把reverse的更新优先。不然可能重复更新。。
这个让我WA苦了。
#include<iostream>
#include<cstdio>
#include<cstdlib>
using namespace std;
#define N 101000
struct tree
{
int left,right;
int lmax,lmin;
int mid;
int len;
int sum;
int br;
}t[4*N];
int n,q;
int small;
int sum;
char str[N];
void build(int cur,int l,int r)
{
t[cur].left=l;
t[cur].br=0;
t[cur].right=r;
t[cur].mid=(l+r)>>1;
t[cur].len=r-l+1;
if (l==r)
{
if (str[l]=='(')
{
t[cur].sum=1;
t[cur].lmax=1;
t[cur].lmin=1;
}
else
{
t[cur].sum=-1;
t[cur].lmax=-1;
t[cur].lmin=-1;
}
}
else
{
int L=cur<<1,R=cur<<1|1;
build(L,l,t[cur].mid);
build(R,t[cur].mid+1,r);
t[cur].sum=t[L].sum+t[R].sum;
t[cur].lmin=min(t[L].lmin,t[R].lmin+t[L].sum);
t[cur].lmax=max(t[L].lmax,t[R].lmax+t[L].sum);
}
}
void update(int cur,int l,int r,int v)
{
if (l==t[cur].left && r==t[cur].right)
{
if (v==1)
{
t[cur].lmax=t[cur].sum=t[cur].len;
t[cur].lmin=1;
}
else
{
t[cur].lmin=t[cur].sum=-t[cur].len;
t[cur].lmax=-1;
}
return ;
}
int L=cur<<1,R=cur<<1|1;
if (t[cur].br&1)
{
t[cur].br++;
t[L].br++;
t[R].br++;
t[L].sum=-t[L].sum;
t[R].sum=-t[R].sum;
int temp=-t[L].lmin;
t[L].lmin=-t[L].lmax;
t[L].lmax=temp;
temp=-t[R].lmin;
t[R].lmin=-t[R].lmax;
t[R].lmax=temp;
}
if (t[cur].len==t[cur].sum)
{
t[L].lmax=t[L].sum=t[L].len;
t[L].lmin=1;
t[R].lmax=t[R].sum=t[R].len;
t[R].lmin=1;
}
if (t[cur].len==-t[cur].sum)
{
t[L].lmin=t[L].sum=-t[L].len;
t[R].lmin=t[R].sum=-t[R].len;
t[L].lmax=-1;
t[R].lmax=-1;
}
int m=t[cur].mid;
if (r<=m)
update(L,l,r,v);
else if (l>m)
update(R,l,r,v);
else
{
update(L,l,m,v);
update(R,m+1,r,v);
}
t[cur].sum=t[L].sum+t[R].sum;
t[cur].lmin=min(t[L].lmin,t[R].lmin+t[L].sum);
t[cur].lmax=max(t[L].lmax,t[R].lmax+t[L].sum);
}
void reverse(int cur,int l,int r)
{
// cout<<cur<<' '<<t[cur].left<<' '<<t[cur].right<<' '<<t[cur].sum<<' '<<t[cur].lmin<<' '<<t[cur].lmax<<' '<<l<<' '<<r
// <<' '<<t[cur].br<<' '<<t[cur].len<<endl;
if (l==t[cur].left && r==t[cur].right)
{
t[cur].sum=-t[cur].sum;
int tt=-t[cur].lmax;
t[cur].lmax=-t[cur].lmin;
t[cur].lmin=tt;
t[cur].br++;
return ;
}
int L=cur<<1,R=cur<<1|1;
if (t[cur].br&1)
{
t[cur].br++;
t[L].br++;
t[R].br++;
t[L].sum=-t[L].sum;
t[R].sum=-t[R].sum;
int temp=-t[L].lmin;
t[L].lmin=-t[L].lmax;
t[L].lmax=temp;
temp=-t[R].lmin;
t[R].lmin=-t[R].lmax;
t[R].lmax=temp;
}
if (t[cur].len==t[cur].sum)
{
t[L].lmax=t[L].sum=t[L].len;
t[L].lmin=1;
t[R].lmax=t[R].sum=t[R].len;
t[R].lmin=1;
}
if (t[cur].len==-t[cur].sum)
{
t[L].lmin=t[L].sum=-t[L].len;
t[R].lmin=t[R].sum=-t[R].len;
t[L].lmax=-1;
t[R].lmax=-1;
}
int m=t[cur].mid;
if (r<=m)
reverse(L,l,r);
else if (l>m)
reverse(R,l,r);
else
{
reverse(L,l,m);
reverse(R,m+1,r);
}
t[cur].sum=t[L].sum+t[R].sum;
t[cur].lmin=min(t[L].lmin,t[R].lmin+t[L].sum);
t[cur].lmax=max(t[L].lmax,t[R].lmax+t[L].sum);
}
void qurey(int cur,int l,int r)
{
// cout<<cur<<' '<<t[cur].left<<' '<<t[cur].right<<' '<<t[cur].sum<<' '<<t[cur].lmin<<' '<<t[cur].lmax<<' '<<l<<' '<<r
// <<' '<<t[cur].br<<' '<<t[cur].len<<endl;
if (l==t[cur].left && r==t[cur].right)
{
small=min(small,sum+t[cur].lmin);
sum+=t[cur].sum;
return ;
}
int L=cur<<1,R=cur<<1|1;
int m=t[cur].mid;
if (t[cur].br&1)
{
t[cur].br++;
t[L].br++;
t[R].br++;
t[L].sum=-t[L].sum;
t[R].sum=-t[R].sum;
int temp=-t[L].lmin;
t[L].lmin=-t[L].lmax;
t[L].lmax=temp;
temp=-t[R].lmin;
t[R].lmin=-t[R].lmax;
t[R].lmax=temp;
}
if (t[cur].len==t[cur].sum)
{
t[L].lmax=t[L].sum=t[L].len;
t[L].lmin=1;
t[R].lmax=t[R].sum=t[R].len;
t[R].lmin=1;
}
if (t[cur].len==-t[cur].sum)
{
t[L].lmin=t[L].sum=-t[L].len;
t[R].lmin=t[R].sum=-t[R].len;
t[L].lmax=-1;
t[R].lmax=-1;
}
if (r<=m)
qurey(L,l,r);
else if (l>m)
qurey(R,l,r);
else
{
qurey(L,l,m);
qurey(R,m+1,r);
}
}
int main()
{
freopen("in.txt","r",stdin);
int i,j,k;
int nc;
char s[33];
scanf("%d",&nc);
int num=1;
while (nc--)
{
scanf("%d%s",&n,str);
scanf("%d",&q);
printf("Case %d:\n",num++);
build(1,0,n-1);
while (q--)
{
scanf("%s",s);
scanf("%d%d",&i,&j);
// cout<<s<<' '<<i<<' '<<j<<endl;
if (s[0]=='q')
{
if ((j-i)&1)
{
sum=small=0;
qurey(1,i,j);
// cout<<sum<<' '<<small<<endl;
if (sum==0 && small>=0)
printf("YES\n");
else
printf("NO\n");
}
else
printf("NO\n");
}
else if (s[0]=='s')
{
scanf("%s",s);
if (s[0]=='(')
update(1,i,j,1);
else
update(1,i,j,-1);
}
else
{
reverse(1,i,j);
}
}
printf("\n");
}
return 0;
}
Bracket Sequence
Time Limit: 3000 ms Memory Limit: 65536 kB Solved: 36 Tried: 311
Submit
Status
Best Solution
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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".
