POJ 1141 Brackets Sequence-白红宇

POJ 1141 Brackets Sequence

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发布时间：2019-06-21

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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 给一个只有括号的字符串，请添加尽量少的括号使其变为合法的，并输出。 合法的定义如下： 1.空序列是合法的。 2.如果s是合法的，那么(s)和[s]也是合法的。 3.如果A和B都是合法的，那么AB也是合法的。 分析 设串S至少需要增加d(S)个括号，转移如下： 1.如果S形如(S`)或者[S`]，转移到d(S`)。 2.如果S至少有两个字符，则可以分为AB，转移到d(A)+d(B)。 边界是：当S为空串，d(S)=0；为单字符时，d(S)=1。 注意：不管S是否满足第一种情况，都需要尝试第二种情况，否则"[][]"会转移到"]["。 由此，定义dp[i][j]为字符串第i个字符到第j个字符变成合法的至少需要添加几个括号。 另外，本题还要输出构造的合法串，于是用个二维数组记录切割的位置，最后递归输出。

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                using namespace std;typedef long long LL;const int maxn = 1e4+5;const int mod = 77200211+233;typedef pair
                
                  pii;#define X first#define Y second#define pb push_back//#define mp make_pair#define ms(a,b) memset(a,b,sizeof(a))const int inf = 0x3f3f3f3f;#define lson l,m,2*rt#define rson m+1,r,2*rt+1typedef long long ll;#define N 100010//dp[i][j]表示区间i到j需要的最少括号数int dp[105][105],pos[105][105];char s[105];bool match(char a,char b){ if((a=='('&&b==')') || (a=='['&&b==']')) return true; return false;}void Print(int i,int j){ if(i>j) return; if(i==j){ if(s[i]=='('||s[i]==')') printf("()"); else printf("[]"); return; } if(pos[i][j]==-1){ if(s[i]=='('){ printf("("); Print(i+1,j-1); printf(")"); }else{ printf("["); Print(i+1,j-1); printf("]"); } }else{ Print(i,pos[i][j]); Print(pos[i][j]+1,j); }}int main(){#ifdef LOCAL freopen("in.txt","r",stdin);#endif // LOCAL while(gets(s+1)){ int len = strlen(s+1);// cout<
                 
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转载于:https://www.cnblogs.com/fht-litost/p/8858178.html

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