POJ 1141 Brackets Sequence

Brackets Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 21978 Accepted: 6166 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

Northeastern Europe 2001

 

题意：

给一组小括号与中括号的序列，添加最少的字符，使该序列变为合法序列，输出该合法序列。

 

代码：

#include<stdio.h>#include<stdlib.h>#include<string.h>#define MAX 300char str[MAX];int dp[MAX][MAX],path[MAX][MAX];         //dp[i][j]用于记录字符串中第s位到第e位变成合法序列需添加多少个字符                                                //path[i][j]用于记录字符串中第s位到第e位变成合法序列的改变路径int len;void init(){int i;memset(dp,-1,sizeof(dp));memset(path,-1,sizeof(path));for(i=0;i<len;i++){dp[i][i]=1;dp[i+1][i]=0;         }}bool match(char a,char b){if(a=='(' && b==')' || a=='[' && b==']') return true;else return false;}void output(int s,int e){if(s>e) return;else if(s==e){if(str[s]=='(' || str[s]==')') printf("()");else printf("[]");return;}else if(path[s][e]==-1){printf("%c",str[s]);output(s+1,e-1);printf("%c",str[e]);return;}else{output(s,path[s][e]);output(path[s][e]+1,e);}}int main(){int i,j,s,e;while(gets(str)!=NULL){len=strlen(str);init();for(i=1;i<len;i++){for(j=0;j<len-i;j++){s=j; e=j+i;if(match(str[s],str[e]))dp[s][e]=dp[s+1][e-1];                  //将dp[i+1][i]初始化为0的原因for(int k=s;k<e;k++){if(dp[s][e]>dp[s][k]+dp[k+1][e] || dp[s][e]==-1){dp[s][e]=dp[s][k]+dp[k+1][e];path[s][e]=k;}}}}output(0,len-1);printf("\n");}return 0;}

 

思路：

dp[i][j]用于记录字符串中第s位到第e位变成合法序列需添加多少个字符

path[i][j]用于记录字符串中第s位到第e位变成合法序列的改变路径

状态转移方程：若str[s]与str[e]可搭配，则dp[s][e]=dp[s+1][e-1];否则，dp[s][e]=dp[s][k]+dp[k+1][e];(k从s到e)