简单的区间DP。。。
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 2518 | Accepted: 1300 |
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
Source
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
char str[220];
int dp[220][220];
int main()
{
while(scanf("%s",str)!=EOF)
{
if(str[0]==‘e‘&&str[1]==‘n‘&&str[2]==‘d‘) break;
int n=strlen(str);
memset(dp,0,sizeof(dp));
for(int len=2;len<=n;len++)
{
for(int i=0;i+len-1<n;i++)
{
int j=i+len-1;
if((str[i]==‘(‘&&str[j]==‘)‘)||(str[i]==‘[‘&&str[j]==‘]‘))
{
dp[i][j]=dp[i+1][j-1]+1;
}
dp[i][j]=max(max(dp[i+1][j],dp[i][j-1]),dp[i][j]);
for(int k=i;k<=j;k++)
dp[i][j]=max(dp[i][k]+dp[k+1][j],dp[i][j]);
}
}
printf("%d\n",2*dp[0][n-1]);
}
return 0;
}
POJ 2955 Brackets,布布扣,bubuko.com
原文：http://blog.csdn.net/ck_boss/article/details/22943925