URAL1183——DFS+回溯—— Brackets Sequence

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 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

([(]

()[()]

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;
}