题解 P1944 最长括号匹配_NOI导刊2009提高(1)


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 ;
}
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