Round 1 F

题目链接：
https://vjudge.net/contest/168327#problem/F

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 SS, and he wants Rikka to choose two different position i,ji,j and swap Si,SjSi,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
For the second sample input, Rikka can choose (1,3) or (2,4) to swap. But do nothing is not allowed.

大意：
括号匹配 进阶版，必须交换一次位置。

法一，开数组记录

#include<bits/stdc++.h>#define mem(s,t) memset(s,t,sizeof(s))typedef long long ll;using namespace std;//#define LOCALconst int MAXN =100000+10;int op[MAXN];int main(){#ifdef LOCAL    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);#endif    int t;    scanf("%d",&t);    while(t--){        int n;        scanf("%d",&n);        char s[MAXN];        mem(s,0);        mem(op,0);        scanf("%s",s+1);        for(int i=1;i<=n;i++){            if(s[i]=='(')                op[i]=op[i-1]+1;            else op[i]=op[i-1]-1;        }        if(op[n]!=0){            puts("No");            continue;        }        if(n%2){            puts("No");            continue;        }        if(strcmp(s+1,"()")==0){            puts("No");            continue;        }        int bad=0;        for(int i=1;i<=n;i++){            if(op[i]<-2)                bad=1;        }        if(bad) puts("No");        else puts("Yes");    }    return 0;}

法二，栈

#include<bits/stdc++.h>#define mem(s,t) memset(s,t,sizeof(s))typedef long long ll;using namespace std;//#define LOCALconst int MAXN =100000+10;stack<char> st;char s[MAXN];int n;bool solve(){    while(!st.empty()) st.pop();    st.push(s[1]);    for(int i=2;i<=n;i++){        if(!st.empty() && st.top()=='(' && s[i]==')')            st.pop();        else st.push(s[i]);    }    if(st.size()>4) return 0;    return 1;}int main(){#ifdef LOCAL    freopen("in.txt","r",stdin);    freopen("out.txt","w",stdout);#endif    int t;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        mem(s,0);        scanf("%s",s+1);        if(n&1) puts("No");        else if(strcmp(s+1,"()")==0) puts("No");        else if(solve()) puts("Yes");        else puts("No");    }    return 0;}