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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 3615 | Accepted: 1874 |
Description
We give the following inductive definition of 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
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<string>
#include<algorithm>
using namespace std;
char s[105];
int maxx,dp[105][105];
int main()
{
while(scanf("%s",s)!=EOF)
{
if(strcmp(s,"end")==0)
break;
memset(dp,0,sizeof(dp));
maxx=0;
int len=strlen(s);
for(int i=0;i<len;i++)
{
for(int j=0,k=i;k<len;j++,k++)
{
if((s[j]==‘(‘&&s[k]==‘)‘||(s[j]==‘[‘&&s[k]==‘]‘)))
dp[j][k]=dp[j+1][k-1]+2;
for(int t=j+1;t<k;t++)
{
if(dp[j][t]+dp[t][k]>dp[j][k])
dp[j][k]=dp[j][t]+dp[t][k];
}
if(dp[j][k]>maxx)
maxx=dp[j][k];
}
}
printf("%d\n",maxx);
}
return 0;
}
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原文地址：http://www.cnblogs.com/a972290869/p/4342626.html
