题目链接
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
题意:给了一个括号序列(只有"(" ")" "[" "]") 现在让添加括号,使括号序列变得匹配,要求添加最少的括号,输出这个匹配的括号序列;
思路:区间DP,dp[i][j]表示区间i~j匹配添加括号后区间最小长度,dp[i][j]=dp[i][k]+dp[k+1][j] ,注意当s[i]=='('&&s[j]==')' || s[i]=='['&&s[j]==']' 时,特判一下dp[i][j]=min(dp[i][j],dp[i+1][j-1]+2); 这样可以找出匹配后的序列最小长度,但是题目要求输出匹配的序列,那么可以在定义一个数组v[i][j] 标记i~j区间的断开位置,如果s[i]=='('&&s[j]==')' || s[i]=='['&&s[j]==']' 时 v[i][j]==-1, 然后在递归调用输出即可;
代码如下:
#include#include #include #include #include using namespace std;const int inf=0x3f3f3f3f;char s[105];int v[105][105];int dp[105][105];void print(int l,int r){ if(r dp[i][k]+dp[k+1][i+l]) { dp[i][i+l]=dp[i][k]+dp[k+1][i+l]; v[i][i+l]=k; } } if(s[i]=='('&&s[i+l]==')'||s[i]=='['&&s[i+l]==']') { if(dp[i][i+l]>dp[i+1][i+l-1]+2) { dp[i][i+l]=dp[i+1][i+l-1]+2; v[i][i+l]=-1; } } } } print(0,len-1); printf("\n"); return 0;}