Brackets Sequence

Time limit 1000 ms Memory limit 65536 kB

区间DP

来源：
poj： poj 1141 Brackets Sequence
vjudge：c-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

()[()]

题意：

输入一串只有 '(' ')' '[' ']'的字符串，求最少加入多少个括号能够使它组成完整。输出完整的字符串

解题思路：

区间DP......
记得自己推一遍，，，你就明白了@_@
你自己是想不到的-_-||

完整代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define inf 0x3ffffff
char s[110];
int dp[110][110],vis[110][110];
void print(int i,int j)
{
    if(i>j)
        return;
    if(i==j)
    {
        if(s[i]=='('||s[i]==')')
          printf("()");
        else if(s[i]=='['||s[i]==']')
          printf("[]");
    }
    else if(vis[i][j]==-1)
    {
        printf("%c",s[i]);
        print(i+1,j-1);
        printf("%c",s[j]);
    }
    else
    {
        print(i,vis[i][j]);
        print(vis[i][j]+1,j);
    }
}
int main()
{
    gets(s);
    int len=strlen(s);
    for(int i=0;i<len;i++)
        dp[i][i]=1;
    for(int l=1;l<len;l++)
        for(int i=0;i<len-l;i++)
        {
           int j=l+i;
           dp[i][j]=inf;
           if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))
            {
                 if(dp[i][j]>dp[i+1][j-1])
                 {
                     dp[i][j]=dp[i+1][j-1];
                     vis[i][j]=-1;///表示s[i] 与s[j] 匹配时是最优的;
                 }
            }
            for(int k=i;k<j;k++)
            {
                 if(dp[i][j]>dp[i][k]+dp[k+1][j])
                 {
                      dp[i][j]=dp[i][k]+dp[k+1][j];
                      vis[i][j]=k;///表示以k为分割点匹配时最优;
                 }
            }
        }
    print(0,len-1);
    printf("\n");
    return 0;
}