一題DP題
輸入一串由'(',')','[',']'組成的字符串,對於[ ]，( )，[ ( ) ], ( [ ] ),都是合法的匹配。而對於( ],[ )等則是不合法的匹配，你的任務是找出括號集中最長合法匹配
我們先處理只有'('和')'的情況,遇到一個')',如果前面是一個'(',那麼長度是2的"()"肯定是可以的了,但是如果'('前面已經有匹配了的,如前面如果是一個已經匹配了的"()",那麼我們需要把它也加進去.
如上,我們再把'['和']'加進來
/*
0.13s DP
*/
#include "stdio.h"#include <string.h>char str[100002] ;int DP[100002] ;bool mark[100002] ;int main ( void ){ int i, j ;
while ( 1 == scanf ( "%s", str ) ) {
int start = 0, end = -1, len = -1 ; DP[0] = -1 ; mark[0] = false ; int pos = strlen(str) ; for ( i = 1 ; i < pos ; i++ ) { if ( str[i] == '(' || str[i] == '[' ) { mark[i] = false ; //DP[i] = -1 ;
} else if ( str[i] == ')' ) { if ( str[i-1] == '(' ) { if ( i-2 >= 0 && mark[i-2] ) { mark[i] = true ; DP[i] = DP[i-2] ; } else { mark[i] = true ; DP[i] = i-1 ; } if ( len < i-DP[i] ) { len = i - DP[i] ; start = DP[i] ; end = i ; } } else if ( mark[i-1] && DP[i-1] != 0 ) { if ( str[DP[i-1]-1] == '(' ) { mark[i] = true ; if ( DP[i-1]-2 >= 0 && mark[DP[i-1]-2] ) DP[i] = DP[DP[i-1]-2] ; else DP[i] = DP[i-1]-1 ; if ( len < i-DP[i] ) { len = i - DP[i] ; start = DP[i] ; end = i ; } } else mark[i] = false ; } else mark[i] = false ; } else if ( str[i] == ']' ) { if ( str[i-1] == '[' ) { if ( i-2 >= 0 && mark[i-2] ) { mark[i] = true ; DP[i] = DP[i-2] ; } else { mark[i] = true ; DP[i] = i-1 ; } if ( len < i-DP[i] ) { len = i - DP[i] ; start = DP[i] ; end = i ; } } else if ( mark[i-1] && DP[i-1] != 0 ) { if ( str[DP[i-1]-1] == '[' ) { mark[i] = true ; if ( DP[i-1]-2 >= 0 && mark[DP[i-1]-2] ) DP[i] = DP[DP[i-1]-2] ; else DP[i] = DP[i-1]-1 ; if ( len < i-DP[i] ) { len = i - DP[i] ; start = DP[i] ; end = i ; } } else mark[i] = false ; } else mark[i] = false ; } } for ( i = start ; i <= end ; i++ ) { printf ( "%c", str[i] ) ; } printf ( " " ) ; } return 0 ;}