POJ2955:Brackets(区间DP)
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
题意:求出互相匹配的括号的总数
思路:一道区间DP,dp[i][j]存的是i~j区间内匹配的个数
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std; int check(char a,char b)
{
if(a=='(' && b==')')
return 1;
if(a=='[' && b==']')
return 1;
return 0;
} int main()
{
char str[105];
int dp[105][105],i,j,k,len;
while(~scanf("%s",str))
{
if(!strcmp(str,"end"))
break;
len = strlen(str);
for(i = 0; i<len; i++)
{
dp[i][i] = 0;
if(check(str[i],str[i+1]))
dp[i][i+1] = 2;
else
dp[i][i+1] = 0;
}
for(k = 3; k<=len; k++)
{
for(i = 0; i+k-1<len; i++)
{
dp[i][i+k-1] = 0;
if(check(str[i],str[i+k-1]))
dp[i][i+k-1] = dp[i+1][i+k-2]+2;
for(j = i; j<i+k-1; j++)
dp[i][i+k-1] = max(dp[i][i+k-1],dp[i][j]+dp[j+1][i+k-1]);
}
}
printf("%d\n",dp[0][len-1]);
} return 0;
}
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