ZOJ2704（Brackets）

Brackets

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Time Limit: 1 Second      Memory Limit: 32768 KB

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Given a string consisting of brackets of two types find its longest substring that is a regular brackets sequence.

Input

There are mutiple cases in the input file.

Each case contains a string containing only characters ‘(’ , ‘)’ , ‘[’ and ‘]’ . The length of the string does not exceed 100,000.

There is an empty line after each case.

Output

Output the longest substring of the given string that is a regular brackets sequence.

There should be am empty line after each case.

Sample Input

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

Sample Output

[()]

//2009-05-11 18:21:20 Accepted  2704 C++ 60 776

#include <iostream>
#include <stack>
#include <cstdio>
#include <cstring>
using namespace std;

const int N = 100004;


typedef struct
{
    char data;
    int pos;
}node;


char str[N];
bool pp[N];

bool isok(char a, char b)
{
    if(a == '(' && b == ')')
        return true;
    if(a == '[' && b== ']')
        return true;
    return false;
}

void solve()
{
    stack<node> S;
    node a;
    int len = strlen(str);
    int i , j, k, s, e;
    int maxp, tmp;
    
    /**/////////////////////////////////

    memset(pp, false, sizeof(pp));
    
    for(i = 0; i < len; i++)
    {
        if(str[i] == '(' || str[i] == '[')
        {//对于左括号，入栈
            a.data = str[i]; a.pos = i;
            S.push(a);
        }
        else 
        {
            if(S.size())
            {//栈非空情况
                a = S.top();
                if(isok(a.data, str[i]))
                {//匹配情况
                    S.pop();
                    pp[a.pos] = true;
                    pp[i] = true;
                }
                else
                {
                    a.data = str[i];
                    a.pos = i;
                    S.push(a);
                }
            }
            else
            {
                a.data = str[i];
                a.pos = i;
                S.push(a);
            }

        }
    }

    maxp = 0;

    i = 0;
    while(i < len)
    {
        while(pp[i] == false && i < len)
            i++;

        if(i >= len)
            break;

        j = i; tmp = 0;
        while(pp[i] && i < len)
            i++;
        k = i - 1;

        if(k - j > maxp)
        {
            maxp = k - j;
            s = j; e = k;
        }
    }

    if(maxp)
    {
        while(s <= e)
            printf("%c", str[s++]);
    }

    printf("\n\n");
    
}
int main()
{
    while(scanf("%s", str) != EOF)
    {
        solve();    
    }
    return 0;
}