Brackets （poj 2955）

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4890 Accepted: 2615

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 m such 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

66406

题意：

给出一个由 ( ) [ ] 组成的串，问这个串中有多少个匹配括号。

思路：

用 dp[i][j] 表示 i~j 这个区间内的最大括号匹配数。

如果 s[i] 和 s[j] 匹配，那么很明显有 dp[i][j] = dp[i+1][j-1] + 2;

如果 s[i] 和 s[j] 不匹配，那么我们可以在 i~j 中找一个断点k 将区间分为 i~k 和 k+1~j ，则有

dp[i][j] = max(dp[i][j],dp[i][k] + dp[k+1][j]);

我们可以通过枚举区间长度来进行转移。

#include"iostream"#include"cstring"#include"cstdio"#include"algorithm"using namespace std;char ss[105];int dp[105][105];bool check(int i,int j){    if(ss[i] == '(' && ss[j] == ')') return true;    if(ss[i] == '[' && ss[j] == ']') return true;    return false;}int main(void){    while(~scanf("%s",ss))    {        if(ss[0] == 'e')            break;        int len = strlen(ss);        memset(dp,0,sizeof(dp));        for(int i = 0;i < len-1;i++)        {            if(check(i,i+1)) dp[i][i+1] = 2;        }        for(int k = 3;k <= len;k++)        {            for(int i = 0;i + k - 1 < len;i++)            {                if(check(i,i+k-1))                    dp[i][i+k-1] = dp[i+1][i+k-2] + 2;                for(int j = i;j < i + k - 1;j++)                    dp[i][i+k-1] = max(dp[i][i+k-1],dp[i][j]+dp[j+1][i+k-1]);            }        }        printf("%d\n",dp[0][len-1]);    }    return 0;}