POJ 1141 Brackets Sequence (区间dp 括号匹配 经典题)

Brackets Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 26407 Accepted: 7443 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

题目链接：http://poj.org/problem?id=1141

题目大意：给一些括号，求让它们合法最少要加上多少括号，并输出加上后的结果，合法的定义：
1.空串是合法的
2.若S是合法的，则［S］和（S）均是合法的
3.若A和B是合法的，则AB是合法的

题目分析：令dp[i][j]表示使子序列从i到j合法要加的最小括号数
当子序列长度为1时，dp[i][i] = 1
当子序列长度不为1时两个方案：
1）dp[i] == '(' && dp[j] == ')'或者dp[i] == '[' && dp[j] == ']'说明最外侧已合法，则要加的括号数由里面的子序列决定即dp[i][j] = dp[i + 1][j - 1]
2）枚举分割点，即i <= k < j，dp[i][j] = min(dp[i][k]，dp[k + 1][j])
这样要添加的最少数量就能得到即dp[0][len - 1]，但是题目要输出序列，因此我们还要记录路径，若s[i] == s[j]则path[i][j] = -1，否则path[i][j] = k(分割点)，输出的时候采用递归的方法见程序注释

#include <cstdio>#include <cstring>int const INF = 0xfffffff;int const MAX = 105;int dp[MAX][MAX], path[MAX][MAX];char s[MAX];void Print(int i, int j){    if(i > j) //无效位置        return;    if(i == j)  //遇单个字符输出匹配后的结果    {        if(s[i] == '(' || s[i] == ')')            printf("()");        else            printf("[]");    }    else if(path[i][j] == -1) //若i到j已经匹配，输出左边，递归中间再输出右边    {        printf("%c", s[i]);        Print(i + 1, j - 1);        printf("%c", s[j]);    }    else  //否则，递归输出分割点两边    {        Print(i, path[i][j]);        Print(path[i][j] + 1, j);    }}int main(){    while(gets(s))    {        int n = strlen(s);        if(n == 0)        {            printf("\n");            continue;        }        memset(dp, 0, sizeof(dp));        for(int i = 0; i < n; i++)            dp[i][i] = 1;        for(int l = 1; l < n; l++)        {            for(int i = 0; i < n - l; i++)            {                int j = i + l;                dp[i][j] = INF;                if((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']'))                {                    dp[i][j] = dp[i + 1][j - 1];                    path[i][j] = -1;                }                for(int k = i; k < j; k++)                {                    if(dp[i][j] > dp[i][k] + dp[k + 1][j])                    {                        dp[i][j] = dp[i][k] + dp[k + 1][j];                        path[i][j] = k;                    }                }            }        }        Print(0, n - 1);        printf("\n");    }}