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  • HDU 5831 Rikka with Parenthesis II （栈+模拟）

    Rikka with Parenthesis II

    题目链接：

    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 $S_i,S_j$. 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.

    Source

    2016 Multi-University Training Contest 8
    ##题意： 括号匹配问题：空、XY、(X) 为合法字串. 给出一个字串，是否可以经过且只经过一次交换操作，使得结果串合法. (不能不交换，不能与自己位置交换)
    ##题解： 考虑交换操作：(必须换且只能换一次). ①. 如果原串本身就合法，长度为2时："()"->"No", ")("->"Yes". 长度大于2时一定为"Yes", 因为可以直接交换两个相同的括号. ②. 如果原串非法，那么交换时一定交换的不同的符号(否则没用). 那么符合条件的串一定是把一个 '(' -> ')' 且一个 ')' -> '(' . 那么只考虑变换即可.
    考虑如何判断一个串是否合法的过程： 依次处理字符，若是'('则入栈，若是')'则从栈中弹出一个'('. 若没有'('则不合法. 那么此题就是上述过程的变种，在处理过程中允许两次变换. 由于'('->')'的时机不方便考虑, 这里就只考虑')'->'('. ①. 如果当前是'('，直接入栈. ②. 如果当前是')'，如果栈非空，则弹出一个'('; 如果栈空就把当前的')'变成'('入栈. (标记最多只能变化一次).
    用flag标记是否有将')'变为'('的操作. 结果栈要么为空，要么全是'('. 1. 如果整个字串没有被处理完，那么肯定是"No". 2. 如果flag=0, 那么要求没有'('剩下. 3. 如果flag=1, 那么结果栈中的'('只能是两个. "((" -> "()".
    官方题解： 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).
    ##代码： ``` cpp #include #include #include #include #include #include #include #include #include #include #include #define LL long long #define eps 1e-8 #define maxn 101000 #define mod 100000007 #define inf 0x3f3f3f3f #define mid(a,b) ((a+b)>>1) #define IN freopen("in.txt","r",stdin); using namespace std;

    char str[maxn];
    stack s;

    int main(int argc, char const *argv[])
    {
    //IN;

    int t; cin >> t;
while(t--)
{
    int n; scanf("%d", &n);
    while(!s.empty()) s.pop();
    scanf("%s", str);

    if(n == 2) {
        if(str[0]=='(' && str[1]==')') {
            puts("No");
            continue;
        }
    }

    int i;
    int flag1 = 0;
    for(i=0; i<n; i++) {
        if(str[i] == '(') {
            s.push('(');
        } else {
            if(!s.empty()) s.pop();
            else {
                if(flag1) break;
                flag1 = 1;
                s.push('(');
            }
        }
    }

    if(i == n) {
        if(!flag1) {
            if(s.empty()) puts("Yes");
            else puts("No");
        }
        else {
            if(s.size() != 2) puts("No");
            else puts("Yes");
        }
    }
    else puts("No");
}

return 0;

    }
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  • 原文地址：https://www.cnblogs.com/Sunshine-tcf/p/5761816.html
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