1626 - Brackets sequence——［动态规划］

Posted 2020-07-10 kiraa

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 a₁a₂...a_n is called a subsequence of the string b₁b₂...b_m, if there exist such indices 1 ≤ i₁ < i₂ < ... < i_n ≤ m, that a_j=b_ij for all 1 ≤ j ≤ n.

Input 

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

 

 

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

Output 

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

 

 

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 

1

([(]

Sample Output 

()[()]


紫书上有详解＋代码，就不废话了直接贴代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 using namespace std;
 5 const int maxn = 110;
 6 char s[maxn];
 7 int d[maxn][maxn];
 8 int n;
 9 inline bool match(char a,char b){
10     if((a==‘(‘&&b==‘)‘)||(a==‘[‘&&b==‘]‘)) return true;
11     else return false;
12 }
13 void dp(){
14     for(int i=0;i<n;i++){
15         d[i+1][i]=0;
16         d[i][i]=1;
17     }
18    
19     for(int i=n-2;i>=0;i--){
20         for(int j=i+1;j<n;j++){
21             d[i][j]=maxn;
22             if(match(s[i],s[j]))
23                 d[i][j]=min(d[i][j],d[i+1][j-1]);
24             for(int k=i;k<j;k++){
25                 d[i][j]=min(d[i][j],d[i][k]+d[k+1][j]);
26             }
27         }
28     }
29 }
30 void print(int i,int j){
31     if(i>j) return;
32     if(i==j){
33         if(s[i]==‘(‘||s[i]==‘)‘) printf("()");
34         else printf("[]");
35         return;
36     }
37     int ans=d[i][j];
38     if(match(s[i],s[j])&&ans==d[i+1][j-1]){
39         printf("%c",s[i]);print(i+1,j-1);printf("%c",s[j]);
40         return;
41     }
42     for(int k=i;k<j;k++){
43         if(ans==d[i][k]+d[k+1][j]){
44             print(i,k);print(k+1,j);
45             return;
46         }
47     }
48     
49 }
50 int main(int argc, const char * argv[]) {
51     int T;
52     scanf("%d",&T);
53     getchar();
54     while(T--){
55         getchar();
56         memset(s, 0, sizeof s);
57         char ch;
58         for(int i=0;(ch=getchar())!=‘\n‘;i++){
59             s[i]=ch;
60         }
61         n=strlen(s);
62         
63         dp();
64         print(0,n-1);
65         printf("\n");
66         if(T!=0) printf("\n");
67     }
68     return 0;
69 }