状态压缩- -Brackets

标签： ACM

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题目：

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

解题思路：
两种满足的条件
第一种如果两括号匹配则等于中间的匹配+2
第二种区间的匹配等于区间中的相邻区间匹配之和
将2个匹配到最大个数匹配遍历一遍最后就是答案

还是直接看AC代码容易理解些吧.......

#include <iostream>
#include <string.h>
using namespace std;
int dp[105][105];   //表示从i到j字符串的最大匹配
bool match(char a ,char b)
{
    return ((a=='('&&b==')')||(a=='['&&b==']'));
}
int max(int a,int b)
{
    return (a>b?a:b);
}
int main()
{
    string s;
    int len,i,j,k,t;
    while(cin>>s)
    {
        memset(dp,0,sizeof(dp));
        if(s=="end")
            break;
        len=s.size();
        for(i=1;i<len;i++)  //匹配i+1长度的字符串
            for(j=0,k=i;k<len;j++,k++)
            {
                if(match(s[j],s[k]))       //寻找最大包含匹配
                    dp[j][k]=dp[j+1][k-1]+2;  //如果j到k匹配，则等于两者中间的匹配度加2
                for(t=j;t<k;t++)          //寻找最大相邻匹配
                    dp[j][k]=max(dp[j][k],dp[j][t]+dp[t+1][k]);
            }
        cout<<dp[0][len-1]<<endl;
    }
    return 0;
}