poj 1141 Brackets Sequence(区间DP，求最小，输出路径，较难)-白红宇

poj 1141 Brackets Sequence(区间DP，求最小，输出路径，较难)

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发布时间：2019-05-24

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1、

2、题目大意

给出一个字符串，只包含()[]四种符号，添加最少的符号使得该字符串有序，跟前面做的括号的问题类似，上次是求字符串中有规律的子串的最长长度，这次是求最少添加多少字符使得有序，并且输出最终有序的字符串来

dp[i][j]表示i到j区间内字符有序添加的最小值，

if(s[i]==s[j]) dp[i][j]=dp[i+1][j-1]

else

dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j])

3、AC代码：

#include
   
    #include
    
     #include
     
      using namespace std;#define N 105char s[N];int dp[N][N];int v[N][N];int check(char a,char b){    if(a=='(' && b==')')        return 1;    if(a=='[' && b==']')        return 1;    return 0;}void output(int l,int r){    if(l>r)        return ;    if(l==r)    {        if(s[l]=='(' || s[l]==')')            printf("()");        else            printf("[]");        return ;    }    if(v[l][r]==-1)    {        printf("%c",s[l]);        output(l+1,r-1);        printf("%c",s[r]);        return ;    }    output(l,v[l][r]);    output(v[l][r]+1,r);}int main(){    //while(scanf("%s",s)!=EOF)    //{    //while多次输入wrong answer    scanf("%s",s);    int len=strlen(s);    memset(dp,-1,sizeof(dp));    memset(v,-1,sizeof(v));    for(int i=0; i
      
       dp[j][k]+dp[k+1][e] || dp[j][e]==-1)                {                    v[j][e]=k;                    dp[j][e]=dp[j][k]+dp[k+1][e];                }            }        }    }    output(0,len-1);    printf("\n");    //}    return 0;}

4、题目：

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 23444		Accepted: 6595		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

()[()]

Source