POJ 1141 Brackets Sequence
Posted Silenceneo_xw
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了POJ 1141 Brackets Sequence相关的知识,希望对你有一定的参考价值。
Brackets Sequence
Time Limit: 1000MS | Memory Limit: 65536K | |||
Total Submissions: 30738 | Accepted: 8817 | 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.
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
()[()]
Source
题目大意:输入一个长度不超过100的括号序列,添加尽量少的括号,使其得到一个合法的括号序列,并且输出这个合法的括号序列。如果存在多个解,输出任意一个合法的括号序列即可。关于合法的括号序列的定义,参见我的另一篇博客:http://www.cnblogs.com/Silenceneo-xw/p/5936944.html,在此不再赘述。
解题思路:与POJ 2955 Brackets这题一样,是一个区间DP的题。思想很类似,但有些许差异。设dp[l][r]表示区间[l,r]内至少需要增加的括号数,则状态转移如下:
1、l==r时,dp[l][r]显然为1;
2、l与r匹配时,dp[l][r] = min(dp[l][r], dp[l+1][r-1]);
3、l与r不匹配时,可将区间[l,r]分成两个区间[l,mid]和[mid+1,r],其中l<=mid<r。
注意:不管上述2是否满足,都要去尝试上述3,否则就会出现"[][]"转移到"]["的情况,这样就只能够添加两个括号了。还有一个注意点就是,当l>r时,要更新dp[l][r]=0,否则在输出结果时,类似"[][]"的结果输不出来,因为不符合判断条件。详见代码。
附上AC代码:
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 using namespace std; 5 const int maxn = 105; 6 const int inf = 0x3f3f3f3f; 7 char str[maxn]; 8 int dp[maxn][maxn]; 9 10 bool match(char a, char b){ 11 return (a==\'(\'&&b==\')\') || (a==\'[\'&&b==\']\'); 12 } 13 14 int dfs(int l, int r){ 15 if (l > r) 16 return dp[l][r] = 0; 17 if (l == r) 18 return dp[l][r] = 1; 19 if (dp[l][r] != inf) 20 return dp[l][r]; 21 int ans = dp[l][r]; 22 if (match(str[l], str[r])) 23 ans = min(ans, dfs(l+1, r-1)); 24 for (int i=l; i<r; ++i) 25 ans = min(ans, dfs(l, i)+dfs(i+1, r)); 26 return dp[l][r] = ans; 27 } 28 29 void print(int l, int r){ 30 if (l > r) 31 return ; 32 if (l == r){ 33 if (str[l]==\'(\' || str[r]==\')\') 34 printf("()"); 35 else 36 printf("[]"); 37 return ; 38 } 39 int ans = dp[l][r]; 40 if (match(str[l], str[r]) && ans==dp[l+1][r-1]){ 41 putchar(str[l]); 42 print(l+1, r-1); 43 putchar(str[r]); 44 return ; 45 } 46 for (int i=l; i<r; ++i) 47 if (ans == dp[l][i]+dp[i+1][r]){ 48 print(l, i); 49 print(i+1, r); 50 return ; 51 } 52 } 53 54 int main(){ 55 while (fgets(str, maxn, stdin)){ 56 int n = strlen(str); 57 str[n-1] = \'\\0\'; 58 --n; 59 memset(dp, inf, sizeof(dp)); 60 dfs(0, n-1); 61 print(0, n-1); 62 putchar(\'\\n\'); 63 } 64 return 0; 65 }
以上是关于POJ 1141 Brackets Sequence的主要内容,如果未能解决你的问题,请参考以下文章