POJ 1141 Brackets Sequence（动态规划 / 递归）

Brackets Sequence (POJ 1141)

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 33360 Accepted: 9662 Special Judge
Description

Let us define a regular brackets sequence in the following way:

1. Empty sequence is a regular sequence.
2. If S is a regular sequence, then (S) and [S] are both regular sequences.
3. If A and B are regular sequences, then AB is a regular sequence.

For example, all of the following sequences of characters are regular brackets sequences:

(), [], (()), ([]), ()[], ()[()]

And all of the following character sequences are not:

(, [, ), )(, ([)], ([(]

Some sequence of characters ‘(‘, ‘)’, ‘[‘, and ‘]’ is given. You are to find the shortest possible regular brackets sequence, that contains the given character sequence as a subsequence. Here, a string a1 a2 … an is called a subsequence of the string b1 b2 … bm, if there exist such indices 1 = i1 < i2 < … < in = m, that aj = bij for all 1 = j = n.
Input

The input file contains at most 100 brackets (characters ‘(‘, ‘)’, ‘[’ and ‘]’) that are situated on a single line without any other characters among them.
Output

Write to the output file a single line that contains some regular brackets sequence that has the minimal possible length and contains the given sequence as a subsequence.
Sample Input

([(]
Sample Output

()[()]

第一眼看到这道题直觉得考的是DP，但我没想到要怎么打印。然后参考了别人的博客，才知道用pos[][]来存储最优分离点的坐标。然后递归打印pos分割的串。

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
const int MAXN = 256;
char str[MAXN];
int dp[MAXN][MAXN],pos[MAXN][MAXN];
int l,r;
bool h(char a,char b)
{
    if(a == '(' && b == ')')return true;
    if(a == '[' && b == ']')return true;
    return false;
}
void print(int i, int j)
{
    if(i > j)return ;
    if(i == j)
    {
        if(str[i] == '(' || str[j] == ')')
        {
            printf("()");
        }
        else
        {
            printf("[]");
        }
    }
    else if(pos[i][j] == -1)
    {
        printf("%c",str[i]);
        print(i+1,j-1);
        printf("%c",str[j]);
    }
    else
    {
        print(i,pos[i][j]);
        print(pos[i][j] + 1,j);
    }

}
int main()
{
    while(gets(str))
    {
        int len;
        len = strlen(str);
        memset(dp,0,sizeof(dp));
        for(int i = 0; i < len; i++)
        {
            dp[i][i] = 1;
        }
        int j;
        for(int k = 1; k < len; k++)
        {
            for(int i = 0; i + k < len; i++)
            {
                j = i + k;
                dp[i][j] = 0x7fffffff;
                if(h(str[i],str[j]))
                {
                    dp[i][j] = dp[i+1][j-1];
                    pos[i][j] = -1;
                }
                for(int mid = i; mid <j; mid++)
                {
                    if(dp[i][j] > dp[i][mid] + dp[mid+1][j])
                    {
                        dp[i][j] = dp[i][mid] + dp[mid+1][j];
                        pos[i][j] = mid;
                    }
                }
            }
        }
        print(0,len-1);
        printf("\n");
    }
    return 0;
}

  

 

  
1
  
2
  
3
  
4
  
5
  
6
  
7
  
8
  
9
  
10
  
11
  
12
  
13
  
14
  
15
  
16
  
17
  
18
  
19
  
20
  
21
  
22
  
23
  
24
  
25
  
26
  
27
  
28
  
29
  
30
  
31
  
32
  
33
  
34
  
35
  
36
  
37
  
38
  
39
  
40
  
41
  
42
  
43
  
44
  
45
  
46
  
47
  
48
  
49
  
50
  
51
  
52
  
53
  
54
  
55
  
56
  
57
  
58
  
59
  
60
  
61
  
62
  
63
  
64
  
65
  
66
  
67
  
68
  
69
  
70
  
71
  
72
  
73
  
74
  
75
  
76
  
77
  
78
  
79
  
80
  
81
 

文章来源: blog.csdn.net，作者：爱玲姐姐，版权归原作者所有，如需转载，请联系作者。

原文链接：blog.csdn.net/jal517486222/article/details/79342521