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区间DP [POJ 1141] Brackets Sequence

2024-07-30 08:02:53   230人阅读

Brackets Sequence
 

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

()[()]
区间DP、重点在输出
#include<iostream>#include<cstdio>#include<cstring>using namespace std;#define max(a,b) ((a)>(b)?(a):(b))#define INF 0x7fffffff#define N 110char s[N];int ss[N][N]; //ss[i][j]=k,i到j从k位置分开添加的括号数最少 int dp[N][N]; //dp[i][j]是在i~j区间最多括号匹配数int judge(char c1,char c2){    if(c1==‘(‘ && c2==‘)‘) return 1;    if(c1==‘[‘ && c2==‘]‘) return 1;    return 0;}void print(int i,int j){    if(i>j) return;    else if(i==j)     {        if(s[i]==‘(‘ || s[i]==‘)‘) cout<<"()";        else cout<<"[]";    }    else    {        if(ss[i][j]==-1)        {            cout<<s[i];            print(i+1,j-1);            cout<<s[j];        }        else        {            print(i,ss[i][j]);            print(ss[i][j]+1,j);        }    }}int main(){    int n,i,j,k,len;    gets(s+1);    n=strlen(s+1);    for(i=1;i<=n;i++)    {        dp[i][i]=1;    }    for(len=2;len<=n;len++)    {        for(i=1;i<=n-len+1;i++)        {            j=i+len-1;            dp[i][j]=INF;            for(k=i;k<j;k++)            {                if(dp[i][j]>dp[i][k]+dp[k+1][j])                 {                    ss[i][j]=k;                    dp[i][j]=dp[i][k]+dp[k+1][j];                }            }            if(judge(s[i],s[j]) && dp[i][j]>dp[i+1][j-1])            {                ss[i][j]=-1;                dp[i][j]=dp[i+1][j-1];            }        }    }    //cout<<dp[1][n]<<endl;    print(1,n);    printf("\n");    return 0;}

 

区间DP [POJ 1141] Brackets Sequence


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