◆练习题目◆◇区间动态规划◇ Brackets Sequence

◇区间动态规划◇ Brackets Sequence

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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

()[()]

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题目大意
给出一个括号序列（最长长度100），包含 “[]” 和 “()” 。插入最少的括号，使序列中的括号全部匹配，注意——”[(])” 不认为是匹配的。输出这个序列，题目设置有 Special judge，可输出任何一个答案。

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题目解析
对于任意一个区间DP，它的状态定义都是：

g[x][y]表示在原序列中第x到第y个元素的答案；
或者，g[x][y]表示在原序列中从第x个元素开始的y个元素的答案；

然后便是状态转移方程式，由于数组g无法同时保存字符串，因此只能保存最少添加的括号数量。分2种情况——

1. 当前第x个字符与第y个字符匹配
2. 当前第x个字符与第y个字符不匹配

通过假设：

设A、B都为完全匹配的括号序列，则AB和 (A)、(B)、[A]、[B] 也是完全匹配的括号序列。

我们可以证出针对情况1的 g[x][y] 应该等于 g[x+1][y-1] ，即忽略当前的位置，计算把中间的序列变成完全匹配的括号序列需要多少次。但是这并不意味着遇到情况1时就不考虑情况2，给出一个简单的反例—— “()()”。若只考虑情况1，则针对 “)(“就会得到2，但是正确答案是0。接下来就考虑情况2，也就是 AB 的情况。我们需要把原串变成 AB 的形式，也就是在某一个点分为2份，因此枚举k，分割为 x~k 和 k+1~y 的两个区间。
最后是设置边界，当 x==y 时，也就是只剩下一个字符时(例如”(” 和 “]”)，只能添加一个括号，即返回1。
输出方案则是通过递归，详见程序（不要说作者懒）。

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程序样例

/*Lucky_Glass*/#include<iostream>#include<algorithm>#include<cstdio>using namespace std;string s;bool f[105][105];int g[105][105];int G(int x,int y){    if(x>y) return 0;    if(x==y) return g[x][y]=1;    if(f[x][y]) return g[x][y];    f[x][y]=true;    if((s[x]=='(' && s[y]==')') || (s[x]=='[' && s[y]==']')) g[x][y]=G(x+1,y-1);    else g[x][y]=y-x+5;    for(int i=x;i<y;i++)        g[x][y]=min(g[x][y],G(x,i)+G(i+1,y));    return g[x][y];}void VJudge(int x,int y){    if(x>y) return;    if(x==y)    {        if(s[x]=='(' || s[y]==')') printf("()");        else printf("[]");        return;    }    for(int i=x;i<y;i++)        if(g[x][y]==g[x][i]+g[i+1][y])        {            VJudge(x,i),VJudge(i+1,y);            return ;        }    if(g[x][y]==g[x+1][y-1] && (s[x]=='(' && s[y]==')') || (s[x]=='[' && s[y]==']')) printf("%c",s[x]),VJudge(x+1,y-1),printf("%c",s[y]);}int main(){    cin>>s;    G(0,(int)s.length()-1);    VJudge(0,(int)s.length()-1);    printf("\n");    return 0;}

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The End

Thanks for reading!

-Lucky_Glass