UVA 673 括號的匹配——經典棧的應用

http://vjudge.net/contest/view.action?cid=50788#problem/A

You are given a string consisting of parentheses () and []. A string of this type is said to be correct:

(a)

if it is the empty string

(b)

if A and B are correct, AB is correct,

(c)

if A is correct, (A ) and [A ] is correct.

Write a program that takes a sequence of strings of this type and check their correctness. Your program can assume that the maximum string length is 128.

Input 

The file contains a positive integer n and a sequence of n strings of parentheses () and [], one string a line.

Output 

A sequence of Yes or No on the output file.

Sample Input 

3
([])
(([()])))
([()[]()])()

Sample Output 

Yes
No
Yes

題目大意：給出你一串由[ ] ( )四種字元組成的字串，判斷是否合法的字串。

大體思路：

方法一：

            因為以前做過一道括號的最大匹配的題，我想著如果最大匹配正好和字串的長度是一致的時候我們可以確定該字串是合法的字串，不知道什麼會超時==

/*
==超時的程式碼==
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace std;
char a[130];
int dp[130][130];
int main()
{
    int T;
    scanf("%d%*c",&T);
    while(T--)
    {
        scanf("%s",a+1);
        memset(dp,0,sizeof(dp));
        int n=strlen(a+1);
        if(n%2)
        {
            printf("No\n");
            continue;
        }
       /* for(int i=1;i<=n;i++)
             cout <<a[i];
        cout <<"**"<<endl;*/
        for(int i=n-1; i>=0; i--)
            for(int j=i+1; j<=n; j++)
            {
                dp[i][j]=dp[i+1][j];
                for(int k=i+1; k<=j; k++)
                    if((a[i]=='('&&a[k]==')')||(a[i]=='['&&a[k]==']'))
                        dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k+1][j]+2);
            }
        /*for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=n;j++)
                printf("%d ",dp[i][j]);
            printf("\n");
        }*/
        if(dp[1][n]==n)
            printf("Yes\n");
        else
            printf("No\n");
    }
    return 0;
}

方法二：

   利用棧的特殊進出順序。詳見程式碼

#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;
  
stack<char> s;
char str[130];
  
int main(void)
{
    int t;
    scanf("%d", &t);
    getchar();
    while (t--)
    {
        gets(str);
        int len = strlen(str);
        if (len & 1)
            puts("No");
        else
        {
            if (!s.empty())
                s.pop();
            s.push('0');
            for (int i = 0; i < len; ++i)
            {
                if (str[i] == '(' || str[i] == '[')
                    s.push(str[i]);
                else if (str[i] == ')')
                {
                    if (s.top() == '(')
                        s.pop();
                    else
                    {
                        s.push('1');
                        break;
                    }
                }
                else
                {
                    if (s.top() == '[')
                        s.pop();
                    else
                    {
                        s.push('1');
                        break;
                    }
                }
            }
            puts(s.top() == '0' ? "Yes" : "No");
        }
    }
    return 0;
}