[ACM] POJ 1141 Brackets Sequence (區間動態規劃）

標籤：des   style   http   color   os   io   strong   for   

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.

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

解題思路：

dp[i][j] 代表從i到j位置中至少添加幾個括弧使得括弧匹配，pos[i][j]= -1,說明 str[i]　str[j]是一對匹配的括弧，否則記錄的是從哪個位置把str[i].....str[j]分為兩部分，輸出答案採用遞迴的形式。

代碼：

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;const int maxn=220;const int inf=0x7fffffff;int pos[maxn][maxn];//從i到j在哪裡分開int dp[maxn][maxn];//從i到j至少添幾個符號char str[maxn];int len;void print(int i,int j){    if(i>j)        return ;//遞迴出口    if(i==j)    {        if(str[i]=='('||str[i]==')')            cout<<"()";        else            cout<<"[]";    }    else if(pos[i][j]==-1)//兩邊是對稱的    {        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++)//長度        for(int i=0;i+k<len;i++)//起點        {            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;//暫時讓它等於-1            }            for(int mid=i;mid<j;mid++)//這個必須要執行的。            {                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;}