hdu 5831 Rikka with Parenthesis II 线段树

Rikka with Parenthesis II

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5831

Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Correct parentheses sequences can be defined recursively as follows:
1.The empty string "" is a correct sequence.
2.If "X" and "Y" are correct sequences, then "XY" (the concatenation of X and Y) is a correct sequence.
3.If "X" is a correct sequence, then "(X)" is a correct sequence.
Each correct parentheses sequence can be derived using the above rules.
Examples of correct parentheses sequences include "", "()", "()()()", "(()())", and "(((())))".

Now Yuta has a parentheses sequence S, and he wants Rikka to choose two different position i,j and swap Si,Sj.

Rikka likes correct parentheses sequence. So she wants to know if she can change S to a correct parentheses sequence after this operation.

It is too difficult for Rikka. Can you help her?

Input

The first line contains a number t(1<=t<=1000), the number of the testcases. And there are no more then 10 testcases with n>100

For each testcase, the first line contains an integers n(1<=n<=100000), the length of S. And the second line contains a string of length S which only contains ‘(’ and ‘)’.

Output

For each testcase, print "Yes" or "No" in a line.

Sample Input

3
4
())(
4
()()
6
)))(((

Sample Output

Yes
Yes
No

Hint

题意

给你一个括号序列，你必须交换俩括号位置，问你可以可以使他合法。

题解：

哎呀，好气啊，感觉只有我们队是用线段树去模拟交换的……

最优情况下一定交换第一个右括号和最后一个左括号，交换后判断一下即可。 时间复杂度 O(n)O(n)

代码

#include <bits/stdc++.h>
#define rep(a,b,c) for(int (a)=(b);(a)<=(c);++(a))
#define drep(a,b,c) for(int (a)=(b);(a)>=(c);--(a))
#define pb push_back
#define mp make_pair
#define sf scanf
#define pf printf
#define two(x) (1<<(x))
#define clr(x,y) memset((x),(y),sizeof((x)))
#define dbg(x) cout << #x << "=" << x << endl;
const int mod = 1e9 + 7;
int mul(int x,int y){return 1LL*x*y%mod;}
int qpow(int x , int y){int res=1;while(y){if(y&1) res=mul(res,x) ; y>>=1 ; x=mul(x,x);} return res;}
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}return x*f;}
using namespace std;
const int maxn = 100000 + 150;
int len,prefix[maxn],lstid;
char str[maxn];
struct Sgtree{
    struct node{
        int l , r , lzy , mi ;

        void Update( int v ){
            lzy += v;
            mi += v;
        }

    }tree[maxn << 2];

    void Build( int l , int r , int o ){
        tree[o].l = l , tree[o].r = r , tree[o].lzy = tree[o].mi = 0;
        if( r > l ){
            int mid = l + r >> 1;
            Build( l , mid , o << 1 );
            Build( mid + 1 , r , o << 1 | 1 );
            Maintain( o );
        }else{
            if( l == len ) lstid = o;
            tree[o].mi = prefix[l];
        }
    }

    int query( int x , int o ){
        int l = tree[o].l , r = tree[o].r;
        if( l == r ) return tree[o].mi;
        else{
            int mid = l + r >> 1 , rs ;
            ReleaseLabel(o);
            if( x <= mid ) rs = query( x , o << 1 );
            else rs = query( x , o << 1 | 1 );
            Maintain(o);
            return rs;
        }
    }

    void Maintain( int o ){
        tree[o].mi = min( tree[o << 1].mi , tree[o << 1 | 1 ].mi );
    }

    void ReleaseLabel( int o ){
        if( tree[o].lzy ){
            tree[o << 1].Update( tree[o].lzy );
            tree[o << 1 | 1].Update( tree[o].lzy );
        }
        tree[o].lzy = 0;
    }

    void Modify( int ql , int qr , int y , int o ){
        int l = tree[o].l , r = tree[o].r;
        if( ql <= l && r <= qr ) tree[o].Update( y );
        else{
            int mid = l + r >> 1;
            ReleaseLabel( o );
            if( ql <= mid ) Modify( ql , qr , y , o << 1 );
            if( qr > mid ) Modify( ql , qr , y , o << 1 | 1 );
            Maintain( o );
        }
    }

    int Search_First( int x , int o ){
        int l = tree[o].l , r = tree[o].r;
        if( l == r ){
            if( tree[o].mi < 0 ) return l;
            return -1;
        }
        int mid = l + r >> 1 , rs = -1;
        ReleaseLabel( o );
        if( tree[o << 1].mi < 0 ) rs = Search_First( x , o << 1 );
        if( rs == -1 && x > mid && tree[o << 1 | 1].mi < 0 ) rs = Search_First( x , o << 1 | 1 );
        Maintain( o );
        return rs;
    }

}Sgtree;


bool solve(){
    rep(i,1,len) if(str[i]=='(' && i > 1){
        int go = Sgtree.Search_First( i - 1 , 1 );
        if( go != -1 ){
            Sgtree.Modify( go , i - 1 , 2 , 1 );
            if( Sgtree.tree[1].mi == 0 && Sgtree.query(len,1) == 0 ) return true;
            Sgtree.Modify( go , i - 1 , -2 , 1 );
        }
    }
    return false;
}

int main(int argc,char *argv[]){
    int T=read();
    while(T--){
        len=read();
        sf("%s",str+1);
        int ar = 0 , ok = 1;
        rep(i,1,len){
            prefix[i] = prefix[i - 1];
            if(str[i]=='('){
                ++ prefix[i];
                ++ ar;
            }
            else{
                if( ar == 0 ) ok = 0;
                -- ar;
                -- prefix[i];
            }
        }
        Sgtree.Build(1,len,1);
        if( ar != 0 ) ok = 0;
        if( ok ){
            if( len == 2 ) pf("No\n");
            else pf("Yes\n");
        }else{
            bool result = solve();
            if( result == true ) printf("Yes\n");
            else printf("No\n");
        }
    }
    return 0;
}