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[ACM] POJ 1141 Brackets Sequence (区间动态规划)
Brackets Sequence
Time Limit: 1000MS | Memory Limit: 65536K | |||
Total Submissions: 25087 | Accepted: 7069 | 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
dp[i][j] 代表从i到j位置中至少添加几个括号使得括号匹配,pos[i][j]= -1,说明 str[i] str[j]是一对匹配的括号,否则记录的是从哪个位置把str[i].....str[j]分为两部分,输出答案采用递归的形式。
代码:
#include <iostream> #include <stdio.h> #include <string.h> using namespace std; const int maxn=220; const int inf=0x7fffffff; int pos[maxn][maxn];//从i到j在哪里分开 int dp[maxn][maxn];//从i到j至少添几个符号 char str[maxn]; int len; void print(int i,int j) { if(i>j) return ;//递归出口 if(i==j) { if(str[i]=='('||str[i]==')') cout<<"()"; else cout<<"[]"; } else if(pos[i][j]==-1)//两边是对称的 { cout<<str[i]; print(i+1,j-1); cout<<str[j]; } else//可分割 { print(i,pos[i][j]); print(pos[i][j]+1,j); } } int main() { cin>>str; len=strlen(str); memset(dp,0,sizeof(dp)); for(int i=0;i<len;i++) dp[i][i]=1; for(int k=1;k<len;k++)//长度 for(int i=0;i+k<len;i++)//起点 { int j=i+k; dp[i][j]=inf; if((str[i]=='('&&str[j]==')')||(str[i]=='['&&str[j]==']')) { dp[i][j]=dp[i+1][j-1]; pos[i][j]=-1;//暂时让它等于-1 } for(int mid=i;mid<j;mid++)//这个必须要执行的。 { if(dp[i][j]>(dp[i][mid]+dp[mid+1][j])) { dp[i][j]=dp[i][mid]+dp[mid+1][j]; pos[i][j]=mid; } } } print(0,len-1); cout<<endl; return 0; }
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