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  • UVALive 7454 Parentheses (栈+模拟)

    Parentheses

    题目链接:

    http://acm.hust.edu.cn/vjudge/contest/127401#problem/A

    Description

    http://7xjob4.com1.z0.glb.clouddn.com/9cf04e507e13a41d77bca3a75bb51e19

    Input

    The first line contains an integer T ≤ 10 indicating the number of test cases. The first line of each test case contains an even integer n, 2 ≤ n ≤ 100, indicating the length of P. Next, the second line gives the sequence P.

    Output

    For each test case, output the minimum number of reverse operations that make P well-formed.

    Sample Input

    3 18 (()())))(((((())(( 2 () 8 (()))()(

    Sample Output

    4 0 2
    ##题意: 括号匹配问题,求最少的翻转(左右括号转换)操作使得原串合法.
    ##题解: 栈模拟判断当前串是否合法: 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. 最后的结果栈中肯定全是'(',那么把其中的一半变成')'即可. 与[HDU 5831](http://www.cnblogs.com/Sunshine-tcf/p/5761816.html)类似.
    ##代码: ``` cpp #include #include #include #include #include #include #include #include #include #include #include #define LL long long #define eps 1e-8 #define maxn 1010 #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);
    
        int ans = 0;
        for(int i=0; i<n; i++) {
            if(str[i] == '(') {
                s.push('(');
            } else {
                if(!s.empty()) s.pop();
                else {
                    ans++;
                    s.push('(');
                }
            }
        }
    
        ans += s.size() / 2; 
    
        printf("%d
    ", ans);
    }
    
    return 0;
    

    }

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  • 原文地址:https://www.cnblogs.com/Sunshine-tcf/p/5781394.html
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