poj 1141 Brackets Sequence

Posted 2020-11-03 北极飞企

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 33911		Accepted: 9816		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：

合法的括号：
1）空序列是一个合法的序列。
2）如果S是一个合法的序列，那么（S）和[S]都是合法的序列。
3）如果A和B是合法的序列，那么AB是合法的序列。
例如，下面给出的序列是合法的括号序列：
(),[],(()),([]),()[],()[()],
不合法的：
（,[,),)(,([)],([(]
输出一行，给出包含了给出序列作为子序列的具有最小可能长度的合法括号序列。

思路：
设阶段r为子序列的长度,(1<=r<=n-1)，状态i为当前子序列的首指针，j=r+i为尾指针，若子序列长度为一的时候，dp[i,i]=1;如果r大于1
的时候，若(s(i)=‘[‘&&s(j)==‘]‘)||(s(i)==‘(‘&&s(j)==‘)‘)那s(i)到s(j)需要添加的最少字符数取决于s(i+1)到s(j-1)，即:
dp[i,j]=dp[i+1,j-1]；否则就需要找到一个过渡的中间变量，即dp[i,j]=min(dp[i,k],dp[k=1，j]);

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#define  N 100
#define INF 0x3f3f3f3f
using namespace std;
char str[N];  //初始串
int dp[N][N]; //状态转移方程组
int path[N][N]; //记录路径
void print(int i,int j) //递归输出路径
{
    if(i>j)
        return ;   //无效位置反回
    if(i==j)       //若子序列含一个字符，则对单个括号输出匹配括号对
    {
        if(str[i]==‘[‘||str[i]==‘]‘)
            printf("[]");
        else
            printf("()");
    }
    else if(path[i][j]==-1)
    {
        printf("%c",str[i]);
        print(i+1,j-1);
        printf("%c",str[j]);
    }
    else
    {
        print(i,path[i][j]);
        print(path[i][j]+1,j);
    }
}

int main()
{
    while(gets(str))
    {
        int n=strlen(str);
        if(n==0)//空行跳过
        {
            printf("\n");
            continue;
        }
        memset(dp,0,sizeof(dp));
        memset(path,0,sizeof(path));
        for(int  i=0; i<n; i++)
            dp[i][i]=1;
        for(int d=1; d<n; d++)//枚举子序列长度
        {
            for(int i=0; i<n-d; i++)//枚举子序列首位置
            {
                int j=i+d;//子序列的尾部
                dp[i][j]=INF;  
                if((str[i]==‘(‘&&str[j]==‘)‘)||(str[i]==‘[‘&&str[j]==‘]‘))
                {
                    dp[i][j]=dp[i+1][j-1];
                    path[i][j]=-1;


                }
                for(int k=i; k<j; k++)//枚举中间指针
                {
                    if(dp[i][j]>dp[i][k]+dp[k+1][j])
                    {
                        dp[i][j]=dp[i][k]+dp[k+1][j];
                        path[i][j]=k; //记录路径
                    }
                }
            }
        }
        print(0,n-1);
        printf("\n");
    }
    return 0;
}