poj 2955 Brackets(括号匹配)

Brackets
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7695 Accepted: 4079
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

题意：给字符串。输出当前字符串有多少个匹配的字符

思路：dp[i][j]为从i字符到j字符的最大匹配数
枚举所有的距离，从下图中 ① 到 ⑤ 枚举最大的值输出
f（i 3~n）
f(j 0~i+j-1 )
f(k i~j-1)
地推方程：
dp[i][j]=dp[i][j],dp[i][k]+dp[k+1][j];

#include <cstring>#include <cstdio>#include <iostream>#include <algorithm>using namespace std;#define INF 0x3f3f3f3fchar str[110];int dp[110][110];int match(int i,int j){    if(str[i]=='('&&str[j]==')')        return 1;    if(str[i]=='['&&str[j]==']')        return 1;    return 0;}int main(){    int n,c;    while(scanf("%s",str)>0&&str[0]!='e')    {        //getchar();        n=strlen(str);        //cout<<n<<endl;        //memset(dp,0,sizeof(dp));        for(int i=0; i<n; i++)        {            dp[i][i]=0;            if(match(i,i+1))                dp[i][i+1]=2;            else                dp[i][i+1]=0;        }        for(int j=3; j<=n; j++)        {            for(int i=0; i+j-1<n; i++)            {                dp[i][i+j-1]=0;                if(match(i,i+j-1))                    dp[i][i+j-1]=dp[i+1][i+j-2]+2;                for(int k=i; k<i+j-1; k++)                {                    dp[i][i+j-1]=max(dp[i][i+j-1],dp[i][k]+dp[k+1][i+j-1]);                }            }        }        cout<<dp[0][n-1]<<endl;    }    return 0;}