POJ 2955 Brackets(区间dp)
2017-03-14 23:18
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Description
We give the following inductive definition of 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:
求出满足括号匹配的最长个数
解题思路:
设d[i][j]表示[i,j]满足括号匹配的最长个数,则有
(1)当s[i]与s[j]匹配:
(2)当s[i]与s[j]不匹配:
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:
求出满足括号匹配的最长个数
解题思路:
设d[i][j]表示[i,j]满足括号匹配的最长个数,则有
(1)当s[i]与s[j]匹配:
d[i][j]=d[i-1][j-1]+2;
(2)当s[i]与s[j]不匹配:
d[i][j]=max(d[i][k]+d[k+1][j] (k在i+1与j之间)
#include<iostream> #include<cstdio> #include<cstring> using namespace std; char s[105]; int d[105][105]; int main() { int n; cin>>n; getchar(); while(n--) { gets(s); memset(d,0,sizeof(d)); int len=strlen(s); for(int k=1;k<len;k++) { for(int i=0,j=k;j<len;i++,j++) { if(s[i]=='('&&s[j]==')'||s[i]=='['&&s[j]==']') d[i][j]=d[i+1][j-1]+2; for(int x=i;x<j;x++) d[i][j]=max(d[i][j],d[i][x]+d[x+1][j]); } } cout<<len-d[0][len-1]<<endl; } return 0; }
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