poj1141Brackets Sequence【区间dp+路径记录】

Brackets Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 27420 Accepted: 7765 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

()[()]

题意：给出一个括号串输出将他们匹配后的结果要求添加最少的括号

#include<cstdio>#include<cstdlib>#include<cstring>using namespace std;char str[110];int dp[110][110];//第i-第j最大配对括号数 int pos[110][110];//第i-第j间加括号的位置 void print(int left,int right){if(left>right)return ;if(left==right){if(str[left]=='('||str[left]==')')printf("()");elseprintf("[]");}else{if(pos[left][right]==-1){putchar(str[left]);print(left+1,right-1);putchar(str[right]);}else{print(left,pos[left][right]);print(pos[left][right]+1,right);}}}int main(){int i,j,k,l,len;while(gets(str)){l=strlen(str);memset(dp,0,sizeof(dp));for(len=1;len<l;++len){for(i=0,j=len;j<l;++j,++i){if((str[i]=='('&&str[j]==')')||(str[i]=='['&&str[j]==']')){dp[i][j]=dp[i+1][j-1]+2;pos[i][j]=-1;//括号已配对不加括号 }for(k=i;k<j;++k){if(dp[i][k]+dp[k+1][j]>=dp[i][j]){dp[i][j]=dp[i][k]+dp[k+1][j];pos[i][j]=k;//在k处需要加括号 }}}}print(0,l-1);putchar('\n');}return 0;}