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
設(shè)dp[i,j]為從位置i到位置j需要加入字符的最小次數(shù)，有dp[i,j]=min(dp[i,k]+dp[k+1,j])，其中i<=k<j。特別的當s[i]='[' s[j]=']'或者s[i]='(' s[j]=')'時，dp[i,j]=dp[i+1,j-1]。初始條件為dp[i,i]=1，其中0<=i<len。
#include <iostream>
using namespace std;
const int MAXN = 110;
char str[MAXN];
int dp[MAXN][MAXN],path[MAXN][MAXN];
void output(int i,int j)
{
if(i>j) return;
if(i==j){
if(str[i]=='[' || str[i]==']') printf("[]");
else printf("()");
}
else if(path[i][j]==-1){
printf("%c",str[i]);
output(i+1,j-1);
printf("%c",str[j]);
}
else{
output(i,path[i][j]);
output(path[i][j]+1,j);
}
}
int main(){
int i,j,k,r,n;
while(gets(str)){
n=strlen(str);
if(n==0){
printf("\n");
continue;
}
memset(dp,0,sizeof(dp));
for(i=0;i<n;i++) dp[i][i]=1;
for(r=1;r<n;r++)
for(i=0;i<n-r;i++){
j=i+r;
dp[i][j]=INT_MAX;
if((str[i]=='(' && str[j]==')')||(str[i]=='[' && str[j]==']'))
if(dp[i][j]>dp[i+1][j-1])
dp[i][j]=dp[i+1][j-1],path[i][j]=-1;
for(k=i;k<j;k++)
if(dp[i][j]>dp[i][k]+dp[k+1][j])
dp[i][j]=dp[i][k]+dp[k+1][j],path[i][j]=k;
}
output(0,n-1);
printf("\n");
}
return 0;
}