首页 > 代码库 > poj 1114
poj 1114
Brackets Sequence
Time Limit: 1000MS | Memory Limit: 65536K | |||
Total Submissions: 25989 | Accepted: 7330 | 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
Northeastern Europe 2001
状态:dp[i][j]表示 i 到 j 最多的匹配个数转移方程:dp[i][j] = max(dp[i][j], dp[i][k] + dp[k+1][j]);#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int N = 105;char str[N];int dp[N][N];bool judge(char a, char b){ if(a == ‘(‘ && b == ‘)‘) return true; if(a == ‘[‘ && b == ‘]‘) return true; return false;}int main(){ while(gets(str) != NULL) { if(!strcmp(str, "end")) break; int len = strlen(str); for(int i = 0; i < len; i++) { dp[i][i] = 0; if(judge(str[i], str[i+1])) dp[i][i+1] = 2; else dp[i][i+1] = 0; } for(int k = 2; k < len; k++) //枚举子串的长度 { for(int i = 0; i + k < len; i++) { int r = i + k; dp[i][r] = 0; if(judge(str[i], str[r])) dp[i][r] = dp[i+1][r-1] + 2; for(int j = i; j < r; j++) dp[i][r] = max(dp[i][r], dp[i][j] + dp[j+1][r]); } } printf("%d\n",dp[0][len-1]); } return 0;}
poj 1114
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。