POJ1141Brackets Sequence动态规划DP

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12516		Accepted: 3353		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

状态：

设d[i][j]表示第i个到第j个括号需要加括号的最小值

p[i][j]记录分割位置

然后递归求解

状态转移方程：

if(s[i,j]=='()'||s[i,j]=='[]')

d[i][j]=d[i+1][j-1]

p[i][j]=-1

d[i][j]=min{d[i][k]+d[k+1][j]}

p[i][j]=k

边界：

d[i][i]=1;

代码：

#include<cstdio> #include<cstring> using namespace std; char s[105]; int dp[105][105],p[105][105],n; void out(int x,int y) { if(x>y) return; if(x==y) { if(s[x]=='('||s[x]==')') printf("()"); else printf("[]"); } else if(!p[x][y]) { putchar(s[x]); out(x+1,y-1); putchar(s[y]); } else { out(x,p[x][y]); out(p[x][y]+1,y); } } void DP() { int i,j,k,m; for(i=1;i<=n;i++) dp[i][i]=1; for(m=1;m<n;m++) for(i=1;i<=n-m;i++) { dp[i][j=i+m]=999999999; if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']')) { dp[i][j]=dp[i+1][j-1]; p[i][j]=0; } for(k=i;k<j;k++) if(dp[i][j]>dp[i][k]+dp[k+1][j]) dp[i][j]=dp[i][p[i][j]=k]+dp[k+1][j]; } out(1,n); puts(""); } int main() { while(gets(s+1)) { n=strlen(s+1); DP(); } }