poj2955括号匹配 区间DP

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5424		Accepted: 2909

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

6
6
4
0

6

 

题意:有一串括号，（） 或者 [ ]这样算作匹配，如果s1匹配了，那么(s1) [s1]也算匹配，其他都是不合法的。

输入一串括号，问你匹配的长度是多少。

 

思路:

区间dp。dp[i]j]表示以i开始的长度为j的匹配的个数。转移的时候还需要考虑 j 到 j+i 之间的这一段，设其最大值x，

如果s[j] == s[j+i] ,x += 1.然后就是状态转移的方程 dp[j][i] = max(x,dp[j][k] + dp[k+j][i-k] | k表示<=i的长度);

因为只进行状态转移的话，无法考虑匹配的情况，所以我先处理匹配时候的值。

 

#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<string>
#include<time.h>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1000000001
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int MAXN = 110;
int dp[MAXN][MAXN];
char s[MAXN];
int ok(char k1,char k2)
{
    if(k1 == '(' && k2 == ')')
        return 1;
    if(k1 == '[' && k2 == ']')
        return 1;
    return 0;
}
int main()
{
    while(~scanf("%s",s+1)){
        if(s[1] == 'e')break;
        int len = strlen(s+1);
        int ans = 0;
        memset(dp,0,sizeof(dp));
        for(int i = 2; i <= len; i++){
            for(int j = 1; j <= len - i + 1; j++){
                if(i == 2 && ok(s[j],s[j+i-1])){
                    dp[j][i] = 1;
                }
                else if(i > 2){
                    int p = 0;
                    for(int k = 0; k <= i - 2; k++){
                        p = max(p,dp[j+1][k]+dp[j+k+1][i-k-2]);
                        //cout<<i<<endl;
                        //cout<<dp[j+1][k]<<' '<<j+1<<' '<<k<<' '<<endl;
                        //cout<<dp[j+k+1][i-k-1]<<' '<<j+k+1<<' '<<i-k-1<<' '<<endl;
                    }
                    if(ok(s[j],s[j+i-1])){
                        p++;
                    }
                    dp[j][i] = max(p,dp[j][i]);
                    for(int k = 0; k <= i; k++){
                        dp[j][i] = max(dp[j][i],dp[j][k]+dp[j+k][i-k]);
                    }
                }
            }
        }
        printf("%d\n",dp[1][len] * 2);
    }
    return 0;
}