题解 P1944 最长括号匹配_NOI导刊2009提高（1）

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栈，模拟

• 把每个元素逐个入栈
• 如果和栈顶元素匹配，那么一块弹出去，同时标记这里是可匹配的。
• 取出连续的，最长的可匹配的序列即可。

#include <iostream>
#include <stdio.h>
#include <string.h>
#define re register
#define Clean(X,K) memset(X,K,sizeof(X))
using namespace std ;
const int Maxl = 1000005 ;
string S ;
char A[Maxl] ;
int Ans = 0 , Now = 0 , Top = 1 , M[Maxl] , Lc[Maxl] , From;
int main () {
//  freopen ("P1944.in" , "r" , stdin) ;
    getline (cin , S) ;
    int L = S.length() ;
    Clean(M , 0) , Clean(A , 0);
    for (re int i = 0 ; i < L ; ++ i) {
        if (S[i] == ')' && A[Top] == '(' ) {
            -- Top , M[i] = M[Lc[Top + 1]] = 1 ;
            continue ;
        }
        if (S[i] == ']' && A[Top] == '[' ) {

            -- Top , M[i] = M[Lc[Top + 1]] = 1 ;
            continue ;
        }
        A[++ Top] = S[i] ;
        Lc[Top] = i ;
    }
    for (re int i = 0 ; i < L; ++ i) {
        if (M[i]) ++ Now ;
        else {
            if (Now > Ans) {
                Ans = Now ;
                From = i - 1 ;
            }
            Now = 0 ;
        }
    }
    int St=0;
    for (re int i = From ; i >= 0 ; -- i ) if (!M[i]) {
            St = i + 1;
            break ;
        };
    for (re int i = St ; i <= From ; ++ i)  putchar(S[i]) ;
    fclose (stdin) , fclose (stdout) ;
    return 0 ;
}