URAL1183——DFS+回溯—— Brackets Sequence

Description

Let us define a regular brackets sequence in the following way:
Empty sequence is a regular sequence.

If S is a regular sequence, then (S) and [S] are both regular sequences.

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 a 1a 2...a n is called a subsequence of the string b 1b 2...b m, if there exist such indices 1 ≤ i 1 < i 2 < ... < i n ≤ m, that a j=b ij for all 1 ≤ j ≤ n.

Input

The input contains at most 100 brackets (characters '(', ')', '[' and ']') that are situated on a single line without any other characters among them.

Output

Write 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

input	output
([(]	()[()]

大意：匹配最少的括号并输出

定义dp[i][j] 表示从i到j最少所需要匹配的括号 ，定义pos[i][j] 表示从i到j是否能括号匹配

状态转移方程 dp[x][y] = min(dp[x][y],dp[x][i]+dp[i+1][y])

杰哥代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[110][110];
int pos[110][110];
char s[110];
const int inf = 0x3f3f3f3f;
void dfs(int x,int y)
{
if(dp[x][y] != -1) return ;
if(x > y) {
dp[x][y] = 0;
return ;
}
if(x == y){
dp[x][y] = 1;
return ;
}
dp[x][y] = inf;
if((s[x] == '(' && s[y] == ')')||(s[x] == '[' && s[y] == ']')){
pos[x][y] = -1;
dfs(x+1,y-1);
dp[x][y] = dp[x+1][y-1];
}
for(int i = x; i < y; i++){
dfs(x,i);
dfs(i+1,y);
if(dp[x][i] + dp[i+1][y] < dp[x][y]){
dp[x][y] = dp[x][i] + dp[i+1][y];
pos[x][y] = i;
}
}
}
void print(int x,int y)
{
if(x > y)
return ;
if(x == y){
if(s[x] == '(' || s[x] == ')')
printf("()");
else
printf("[]");
return ;
}
if(pos[x][y] == -1){
printf("%c",s[x]);
print(x+1,y-1);
printf("%c",s[y]);
}

else {
print(x,pos[x][y]);
print(pos[x][y]+1,y);
}
}
int main()
{
scanf("%s",s+1);
int n = strlen(s+1);
memset(dp,-1,sizeof(dp));
dfs(1,n);
print(1,n);
printf("\n");
return 0;
}