Brackets Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 14226 Accepted: 7476 Description
We give the following inductive definition of a “regular brackets” sequence:
the empty sequence is a regular brackets sequence, if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and if a and b are regular brackets sequences, then ab is a regular brackets sequence. no other sequence is a regular brackets sequence For instance, all of the following character sequences are regular brackets sequences:
(), [], (()), ()[], ()[()]
while the following character sequences are not:
(, ], )(, ([)], ([(]
Given a brackets sequence of characters a1a2 … an, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i1, i2, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim is a regular brackets sequence.
Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].
Input
The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (, ), [, and ]; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.
Output
For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.
Sample Input
((())) ()()() ([]]) )[)( ([][][) end Sample Output
6 6 4 0 6 Source
Stanford Local 2004
代码语言：javascript复制#include<algorithm>
#include<iostream>
#include<cmath>
#include<cstring>
#include<cstdio>
using namespace std;
bool match(char a,char b);
#define mst(a,b) memset((a),(b),sizeof(a))
const int maxn=500;
int dp[maxn][maxn];
int main()
{
string ob;
while(cin>>ob)
{
if(ob=="end") break;
mst(dp,0);
for(int len=2;len<=ob.length( );len ){
for(int i=1;i<=ob.length( ) 1-len;i ){
int j=len i-1;
if(match(ob[i-1],ob[j-1])) dp[i][j]=dp[i 1][j-1] 2;
for(int k=i;k<j;k )
{
dp[i][j]=max(dp[i][j],dp[i][k] dp[k 1][j]);
}
}
}
// for(int len=1;len<ob.length( )-2;len ) cout<<dp[len][len 3]<<' ';
cout<<dp[1][ob.length( )]<<endl;
}
}
bool match(char a,char b)
{
if(a=='('&&b==')') return 1;
if(a=='['&&b==']') return 1;
else return 0;
}
