POJ 1141 Brackets Sequence

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 30738		Accepted: 8817		Special Judge

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

Northeastern Europe 2001

 

题目大意：输入一个长度不超过100的括号序列，添加尽量少的括号，使其得到一个合法的括号序列，并且输出这个合法的括号序列。如果存在多个解，输出任意一个合法的括号序列即可。关于合法的括号序列的定义，参见我的另一篇博客：http://www.cnblogs.com/Silenceneo-xw/p/5936944.html，在此不再赘述。

 

解题思路：与POJ 2955 Brackets这题一样，是一个区间DP的题。思想很类似，但有些许差异。设dp[l][r]表示区间[l,r]内至少需要增加的括号数，则状态转移如下：

1、l==r时，dp[l][r]显然为1；

2、l与r匹配时，dp[l][r] = min(dp[l][r], dp[l+1][r-1])；

3、l与r不匹配时，可将区间[l,r]分成两个区间[l,mid]和[mid+1,r]，其中l<=mid<r。

注意：不管上述2是否满足，都要去尝试上述3，否则就会出现"[][]"转移到"]["的情况，这样就只能够添加两个括号了。还有一个注意点就是，当l>r时，要更新dp[l][r]=0，否则在输出结果时，类似"[][]"的结果输不出来，因为不符合判断条件。详见代码。

 

附上AC代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 105;
const int inf = 0x3f3f3f3f;
char str[maxn];
int dp[maxn][maxn];

bool match(char a, char b){
    return (a=='('&&b==')') || (a=='['&&b==']');
}

int dfs(int l, int r){
    if (l > r)
        return dp[l][r] = 0;
    if (l == r)
        return dp[l][r] = 1;
    if (dp[l][r] != inf)
        return dp[l][r];
    int ans = dp[l][r];
    if (match(str[l], str[r]))
        ans = min(ans, dfs(l+1, r-1));
    for (int i=l; i<r; ++i)
        ans = min(ans, dfs(l, i)+dfs(i+1, r));
    return dp[l][r] = ans;
}

void print(int l, int r){
    if (l > r)
        return ;
    if (l == r){
        if (str[l]=='(' || str[r]==')')
            printf("()");
        else
            printf("[]");
        return ;
    }
    int ans = dp[l][r];
    if (match(str[l], str[r]) && ans==dp[l+1][r-1]){
        putchar(str[l]);
        print(l+1, r-1);
        putchar(str[r]);
        return ;
    }
    for (int i=l; i<r; ++i)
        if (ans == dp[l][i]+dp[i+1][r]){
            print(l, i);
            print(i+1, r);
            return ;
        }
}

int main(){
    while (fgets(str, maxn, stdin)){
        int n = strlen(str);
        str[n-1] = '\0';
        --n;
        memset(dp, inf, sizeof(dp));
        dfs(0, n-1);
        print(0, n-1);
        putchar('\n');
    }
    return 0;
}