hrbust 1054

最新推荐文章于 2025-09-11 23:21:03 发布

转载最新推荐文章于 2025-09-11 23:21:03 发布 · 80 阅读

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原文链接：http://www.cnblogs.com/mcgrady_ww/p/7896331.html

文章标签：

#php #java

本文介绍了一个简单的算法，用于检查字符串中的括号是否正确配对。通过使用栈数据结构，该算法能够有效地处理圆括号、方括号和花括号等六种类型的括号。

1、链接

http://acm.hrbust.edu.cn/index.php?m=ProblemSet&a=showProblem&problem_id=1054

2、题目

Description
There are six kinds of brackets: ‘(‘, ‘)’, ‘[‘, ‘]’, ‘{’, ‘}’. dccmx’s girl friend is now learning java programming language, and got mad with brackets! Now give you a string of brackets. Is it valid? For example: “(([{}]))” is valid, but “([)]” is not.
Input
First line contains an integer T (T<=10): the number of test case.
Next T lines, each contains a string: the input expression consists of brackets.
The length of a string is between 1 and 100.

Output
For each test case, output “Valid” in one line if the expression is valid, or “Invalid” if not.
Sample Input
2
{{[[(())]]}}
({[}])
Sample Output
Valid
Invalid

3、分析

遇到左括号类就入栈，当遇到右括号类判断其与栈顶元素是否匹配即可

4、代码

#include<bits/stdc++.h>
using namespace std;


char a[105];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s",&a);
        stack<char>st;
        int len = strlen(a);
        int flag = 0;
        for(int i = 0; i < len ; i++)
        {
            if(a[i] == '(' || a[i] == '[' || a[i] == '{')
                st.push(a[i]);

            else if(a[i] == ')')
            {
                if(!st.empty() && st.top() == '(')
                {
                    flag = 1;
                    st.pop();
                }
                else{
                    flag = 0;
                    break;
               }
            }
            else if(a[i] == ']')
            {
                if(!st.empty() && st.top() == '[')
                {
                    flag = 1;
                    st.pop();
                }
                else{
                    flag = 0;
                    break;
               }
            }

            else if(a[i] == '}')
            {
                if(!st.empty() && st.top() == '{')
                {
                    flag = 1;
                    st.pop();
                }
               else{
                    flag = 0;
                    break;
               }
            }
        }

        if(flag && st.empty())
            printf("Valid\n");
        else
            printf("Invalid\n");
    }

    return 0;
}