（POJ

（POJ - 2955）Brackets

Time Limit: 1000MS Memory Limit: 65536K

Total Submissions: 8871 Accepted: 4757

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

题目大意：求一个括号序列的最长匹配括号序列的长度。

思路：设f[i][j]表示区间[i,j]中的最长合法括号序列，s为输入的字符数组，考虑当前区间最优解的可能构成情况：①s[i],s[j]构成一对括号，那么最优解可能为f[i+1][j-1]+2；②最优解由两个括号序列拼成。枚举中间断点k来考虑所有可能的每一种情况中的最优解为f[i][k]+f[k+1][j]。
所以最后在所有情况中选择最大值得状态转移方程为f[i][j]=max(f[i+1][j−1]+2,f[i][k]+f[k+1][j])。

ps：这种括号的题很可能是区间dp或者普通dp题。

递推写法：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int INF=0x3f3f3f3f;const int maxn=105;char s[maxn];int f[maxn][maxn];int main(){    while(scanf("%s",s+1)!=EOF&&strcmp(s+1,"end"))    {        int len=strlen(s+1);        for(int i=1;i<=len;i++) f[i][i]=0;        for(int l=2;l<=len;l++)            for(int i=1;i+l-1<=len;i++)            {                int L=i,R=i+l-1;                f[L][R]=-INF;                if((s[L]=='('&&s[R]==')')||(s[L]=='['&&s[R]==']'))                    f[L][R]=f[L+1][R-1]+2;                for(int k=L;k<R;k++)                    f[L][R]=max(f[L][R],f[L][k]+f[k+1][R]);            }        printf("%d\n",f[1][len]);    }    return 0;}

记忆化搜索写法：

#include<cstdio>#include<cstring>#include<string>#include<iostream>using namespace std;const int INF=0x3f3f3f3f;const int maxn=205;int f[maxn][maxn];string s;int dfs(int i,int j){    if(f[i][j]!=-1) return f[i][j];    if(i>=j) return 0;    if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))        f[i][j]=dfs(i+1,j-1)+2;    for(int k=i;k<j;k++)        f[i][j]=max(f[i][j],dfs(i,k)+dfs(k+1,j));    return f[i][j];}int main(){    while(cin>>s)    {        if(s=="end") break;        memset(f,-1,sizeof(f));        int len=s.size();        printf("%d\n",dfs(0,len-1));    }    return 0;}