J – 括号匹配
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status
Description
We give the following inductive definition of a “regular brackets” sequence:
- the empty sequence is a regular brackets sequence,
- if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and
- if a and b are regular brackets sequences, then ab is a regular brackets sequence.
- no other sequence is a regular brackets sequence
For instance, all of the following character sequences are regular brackets sequences:
(), [], (()), ()[], ()[()]
while the following character sequences are not:
(, ], )(, ([)], ([(]
Given a brackets sequence of characters a1a2 … an, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i1, i2, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim is a regular brackets sequence.
Given the initial sequence ([([]])]
, the longest regular brackets subsequence is [([])]
.
Input
The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (
, )
, [
, and ]
; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.
Output
For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.
Sample Input
((()))
()()()
([]])
)[)(
([][][)
end
Sample Output
6
6
4
0
6题目大意:就是寻找有多少对括号匹配,并输出它们的个数
解题思路:
两种括号的匹配问题,这个问题不能用顺序解决,因为两种括号存在交叉的情形,之前我们用队列的方法做过,而且有些情况并不能知道优先匹配哪种括号,但是今天不用队列方法,用DP
这个题目注意分段max(d[j][i+j-1],d[j][k]+d[k+1][j+i-1])
程序代码:
#include <cstdio>
#include <cstring>
using namespace std;
char s[];
int d[][];
bool match(char a, char b)
{
return (a == '(' && b == ')') || (a == '[' && b == ']');
}
int max(int x,int y)
{
return x>y?x:y;
}
int main()
{
while(~scanf("%s",s))
{
if(strcmp(s,"end")==) break;
int len=strlen(s);
for(int i=;i<len;i++)
{
d[i][i]=;
if(match(s[i],s[i+]))
d[i][i+]=;
else
d[i][i+]=;
}
for(int i=;i<=len;i++)
for(int j=;j+i-<len;j++)
{
d[j][i+j-]=;
if(match(s[j],s[i+j-]))
d[j][i+j-]=d[j+][i+j-]+;
for(int k=j;k<i+j-;k++)
d[j][i+j-]=max(d[j][i+j-],d[j][k]+d[k+][j+i-]);
}
printf("%d\n",d[][len-]);
}
return ;
}