Uva--1626（动规，路径）

2014-09-06 14:58:25

1626 - Brackets sequence

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 a₁a₂...a_n is called a subsequence of the string b₁b₂...b_m, if there exist such indices 1 ≤ i₁ < i₂ < ... < i_n ≤ m, that a_j=b_ij for all 1 ≤ j ≤ n.

Input 

The input begins with a single positive integer on a line by itself indicating the number of the cases following, each of them as described below. This line is followed by a blank line, and there is also a blank line between two consecutive inputs.

  

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

Output 

For each test case, the output must follow the description below. The outputs of two consecutive cases will be separated by a blank line.

  

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 

1

([(]

Sample Output 

()[()]

思路：小白书例题，dp[i][j]表示匹配区间[i,j]内的字符的最优解。不多说了，注意一个trick：输入可能为空串，所以要用gets（or fgets）函数输入。

/*************************************************************************
    > File Name: Uva1626.cpp
    > Author: Nature
    > Mail: 564374850@qq.com
    > Created Time: Sat 06 Sep 2014 12:41:55 PM CST
************************************************************************/

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
using namespace std;
const int INF = 1 << 29;

int T,len;
int dp[105][105];
char s[105];

bool Match(char a,char b){
    if(a == '(' && b == ')') return true;
    if(a == '[' && b == ']') return true;
    return false;
}

void Print(int x,int y){
    if(x > y) return;
    if(x == y){
        if(s[x] == '(' || s[x] == ')') printf("()");
        else printf("[]");
        return;
    }
    if(Match(s[x],s[y]) && dp[x][y] == dp[x + 1][y - 1]){
        printf("%c",s[x]);
        Print(x + 1,y - 1);
        printf("%c",s[y]);
    }
    else{
        int k;
        for(k = x; k < y; ++k)
            if(dp[x][y] == dp[x][k] + dp[k + 1][y]) break;
        Print(x,k);
        Print(k + 1,y);
    }
}

int main(){
    scanf("%d%*c%*c",&T);
    while(T--){
        gets(s + 1);
        getchar();
        len = strlen(s + 1);
        memset(dp,0,sizeof(dp));
        for(int i = 0; i < 105; ++i) dp[i][i] = 1;
        for(int i = len - 1; i >= 1; --i){
            for(int j = i + 1; j <= len; ++j){
                dp[i][j] = INF;
                if(Match(s[i],s[j]))
                    dp[i][j] = min(dp[i][j],dp[i + 1][j - 1]);
                for(int k = i; k < j; ++k)
                    dp[i][j] = min(dp[i][j],dp[i][k] + dp[k + 1][j]);
            }
        }
        Print(1,len);
        printf("\n");
        if(T) printf("\n");
    }
    return 0;
}