poj -2955 Brackets

http://poj.org/problem?id=2955

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2707		Accepted: 1403

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 … a_n, 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 msuch that for indices i₁, i₂, …, i_m where 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im 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
#include
int dp[100][100];
int max(int x,int y)
{
    if(x>y)
       return x;
    else
      return y;
}
bool match(char x,char y)
{
    if(x=='['&&y==']')
       return true;
    else if(x=='('&&y==')')
       return true;
    else
        return false;
}
int main()
{
      int len,i,j,k,g;
     char str[100];
     while(gets(str))
     {
         if(str[0]=='e')
           break;
        memset(dp,0,sizeof(dp));
         len=strlen(str);
         for(i=0;i

 别人的代码，可供参考。

#include
#include
#include
#include
using namespace std;
const int maxn=1002;
char s[maxn];
int dp[maxn][maxn];
int main()
{
    //freopen("//media/学习/ACM/input.txt","r",stdin);
    while(scanf("%s",s),s[0]!='e')
    {
        int i,j,k,n=strlen(s);
        for(i=0;i=0;i--)
        {
            for(j=i+1;j<=n-1;j++)
            {
                 dp[i][j]=max(dp[i+1][j],dp[i][j-1]);
                 for(k=i+1;k<=j;k++)
                 {
                    if((s[i]=='('&&s[k]==')')||(s[i]=='['&&s[k]==']'))
                        dp[i][k]=max(dp[i][k],dp[i+1][k-1]+2);

                    dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
                 }
               //  cout<

 

转载于:https://www.cnblogs.com/cancangood/p/3721013.html