Let us define a regular brackets sequence in the following way:
- Empty sequence is a regular sequence.
- If S is a regular sequence, then (S) and [S] are both regular sequences.
- 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...an is called a subsequence of the string b1b2...bm, if there exist such indices 1 ≤ i1 < i2 < ... < in ≤ m, that aj=bij for all 1 ≤ j ≤ n.
Input
The input contains at most 100 brackets (characters '(', ')', '[' and ']') that are situated on a single line without any other characters among them.
Output
Write 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 | output |
|---|---|
([(] |
()[()] |
Problem Author: Andrew Stankevich Problem Source: 2001-2002 ACM Northeastern European Regional Programming Contest
Difficulty: 250 Printable version Submit solution Discussion (18) My submissions All submissions (5352) All accepted submissions (1722) Solutions rating (1349)
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题解：本题和石子合并问题和求矩阵代价类似，但在此过程中加了一个dir[][]寻找最优路径。俩字经典。
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1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 #include<cstdio> 5 #include<cmath> 6 #include<cctype> 7 #include<algorithm> 8 #include<stack> 9 #include<queue> 10 #include<vector> 11 using namespace std; 12 int dp[1000][1000],i,j; 13 string str; 14 int n,k,t; 15 int dir[1000][1000]; 16 void init() 17 { 18 cin>>str; 19 getchar(); 20 n=str.length(); 21 for(int i=n;i>=1;i--) 22 str[i]=str[i-1]; 23 //for(i=1;i<=n;i++) 24 //cout<<str[i]; 25 //cout<<endl; 26 for(int i=1;i<=n;i++) 27 dp[i][i]=1; 28 } 29 void solve() 30 { 31 for(t=1;t<n;t++) 32 for(i=1;i<=n-t;i++) 33 { 34 j=i+t; 35 dp[i][j]=99999999; 36 if((str[i]=='('&&str[j]==')')||(str[i]=='['&&str[j]==']')) 37 { 38 dp[i][j]=dp[i+1][j-1]; 39 dir[i][j]=0; 40 } 41 for(int k=i;k<j;k++) 42 if(dp[i][k]+dp[k+1][j]<dp[i][j]) 43 { 44 dp[i][j]=dp[i][k]+dp[k+1][j]; 45 dir[i][j]=k; 46 } 47 48 if(str[i]=='('||str[i]=='[') 49 if(dp[i+1][j]+1<dp[i][j]) 50 { 51 dp[i][j]=dp[i+1][j]+1; 52 dir[i][j]=-1; 53 } 54 if(str[i]==')'||str[i]==']') 55 if(dp[i][j-1]+1<dp[i][j]) 56 { 57 dp[i][j]=dp[i][j-1]+1; 58 dir[i][j]=-2; 59 } 60 61 } 62 } 63 void print(int i,int j) 64 { 65 if(i>j)return; 66 if(i==j) 67 { 68 if(str[i]=='('||str[i]==')') 69 cout<<"()"; 70 else 71 cout<<"[]"; 72 73 } 74 else 75 { 76 if(dir[i][j]==0) 77 { 78 if(str[i]=='('&&str[j]==')') 79 { 80 cout<<'('; 81 print(i+1,j-1); 82 cout<<')'; 83 } 84 if(str[i]=='['&&str[j]==']') 85 { 86 cout<<'['; 87 print(i+1,j-1); 88 cout<<']'; 89 } 90 } 91 if(dir[i][j]>0) 92 { 93 print(i,dir[i][j]); 94 print(dir[i][j]+1,j); 95 } 96 if(dir[i][j]==-1) 97 { 98 if(str[i]=='[') 99 { 100 cout<<str[i]; 101 print(i+1,j); 102 cout<<']'; 103 } 104 if(str[i]=='(') 105 { 106 cout<<str[i]; 107 print(i+1,j); 108 cout<<')'; 109 } 110 } 111 if(dir[i][j]==-2) 112 { 113 if(str[i]==']') 114 { 115 cout<<'['; 116 print(i,j-1); 117 cout<<str[i]; 118 } 119 if(str[i]==')') 120 { 121 cout<<'('; 122 print(i,j-1); 123 cout<<str[i]; 124 } 125 } 126 } 127 } 128 int main() 129 { 130 init(); 131 solve(); 132 print(1,n); 133 return 0; 134 }