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

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;

}