POJ 1141 Brackets Sequence

Brackets Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 32855 Accepted: 9508 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

()[()]

题意：

给你一个括号字符串，你的任务是打印处配对好的括号字符串。

思路：区间dp加路径输出。路径输出比较难想。

dp[i][j] 表示从i到j  所需要配对的最小字符个数。（可以参考POJ 2955 ）

http://blog.csdn.net/yjt9299/article/details/78150774

但是两个dp数组表示的意思正好相反。

这里对于这个题，还需要一个pos数组pos[i][j] 记录断开的位置，（从i到j） 

dp方程： dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);

代码：

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define N 305#define inf 0x3f3f3fusing namespace std;int dp[N][N];int pos[N][N];char s[N];void solve(){int i,j,k,d;memset(dp,0,sizeof(dp));memset(pos,-1,sizeof(pos));int len=strlen(s);for(i = 1; i < len; i++)        dp[i][i-1] = 0;for(i=1;i<len;i++) dp[i][i]=1;for(d=1;d<len;d++){for(i=0;i<len-d;i++){j=d+i;dp[i][j]=inf;if((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']')) dp[i][j]=min(dp[i][j],dp[i+1][j-1]);//i  ，j 字符匹配for(k=i;k<j;k++){if(dp[i][k]+dp[k+1][j]<dp[i][j]){dp[i][j]=dp[i][k]+dp[k+1][j];//找到使得从i到j中需要配对的括号的个数为最小的k，也就是在位置k处进行括号配对。 pos[i][j]=k;}}}}return ;}void print(int l,int r){if(l>r) return ;if(l==r){if(s[l]=='('||s[l]==')') printf("()");else if(s[l]=='['||s[l]==']') printf("[]");return ;}else{if(pos[l][r]!=-1){print(l,pos[l][r]);print(pos[l][r]+1,r);}else{if(s[l]=='('){printf("(");print(++l,--r);printf(")");}else if(s[l]=='['){printf("[");print(++l,--r);printf("]");}}}return ;}int main(){scanf("%s",s);int len=strlen(s);solve();for(int i=0;i<len;i++){for(int j=0;j<len;j++){printf("%d ",pos[i][j]);}printf("\n");}print(0,len-1);printf("\n");return 0;}

在最后粘一下大神的思路：

解题思路：
根据“黑书”的思路，定义：
d[i][j]为输入序列从下标i到下标j最少需要加多少括号才能成为合法序列。0<=i<=j<len (len为输入序列的长度)。
c[i][j]为输入序列从下标i到下标j的断开位置，如果没有断开则为-1。
当i==j时，d[i][j]为1
当s[i]=='(' && s[j]==')' 或者 s[i]=='[' && s[j]==']'时，d[i][j]=d[i+1][j-1]
否则d[i][j]=min{d[i][k]+d[k+1][j]} i<=k<j ，  c[i][j]记录断开的位置k
采用递推方式计算d[i][j]
输出结果时采用递归方式输出print(0, len-1)
输出函数定义为print(int i, int j)，表示输出从下标i到下标j的合法序列
当i>j时，直接返回，不需要输出
当i==j时，d[i][j]为1，至少要加一个括号，如果s[i]为'(' 或者')'，输出"()"，否则输出"[]"
当i>j时，如果c[i][j]>=0，说明从i到j断开了，则递归调用print(i, c[i][j]);和print(c[i][j]+1, j);
                如果c[i][j]<0，说明没有断开，如果s[i]=='(' 则输出'('、 print(i+1, j-1); 和")"
                                                                     否则输出"[" print(i+1, j-1);和"]"
黑书上的半括号的那种情况是多余的，包含在d[i][j]=min{d[i][k]+d[k+1][j]}这种情况里。
这种dp比较训练思维，不是常规的那种一眼就看出来的最优子结构。