POJ 2955 Brackets【区间DP】
2018-02-11 12:31
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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:
求括号的最大匹配数.
分析:
不要太拘泥于写法,构造出第i项和第i +1项之间的关系,找到对应的状态方程,再补充细节部分。
当有匹配字符时:dp[i][j]=dp[i+1][j−1]+2;dp[i][j]=dp[i+1][j−1]+2;
暴力区间合并:dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
i,ji,j分别表示左端和右端,区间转移方程成立,再补充细节部分,即可解决.
有点类似这道题的处理方式:Light OJ 1025;
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:
求括号的最大匹配数.
分析:
不要太拘泥于写法,构造出第i项和第i +1项之间的关系,找到对应的状态方程,再补充细节部分。
当有匹配字符时:dp[i][j]=dp[i+1][j−1]+2;dp[i][j]=dp[i+1][j−1]+2;
暴力区间合并:dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
i,ji,j分别表示左端和右端,区间转移方程成立,再补充细节部分,即可解决.
#include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; typedef long long LL; const LL mod = 1e8; const int MAXN = 1e3 + 10; int dp[MAXN][MAXN]; char str[MAXN]; int main() { while(scanf("%s", str + 1) && str[1] != 'e') { memset(dp, 0, sizeof(dp)); int n = strlen(str + 1); for(int len = 2; len <= n; len++) { for(int i = 1; i <= n - len + 1; i++) { int j = i + len - 1; if(str[i] == '[' && str[j] == ']' || str[i] == '(' && str[j] == ')') { dp[i][j] = dp[i + 1][j - 1] + 2; } for(int k = i; k < j; k++) { dp[i][j] = max(dp[i][j], dp[i][k] + dp[k + 1][j]); } } } printf("%d\n", dp[1] ); } return 0; }
有点类似这道题的处理方式:Light OJ 1025;
#include <cstdio> #include <algorithm> #include <cstring> #include <cmath> using namespace std; typedef long long LL; const LL mod = 1e8; const int MAXN = 1e3 + 10; int dp[MAXN][MAXN]; char str[MAXN]; int main() { while(scanf("%s", str + 1) && str[1] != 'e') { memset(dp, 0, sizeof(dp)); int n = strlen(str + 1); for(int len = 2; len <= n; len++) { for(int i = 1; i <= n - len + 1; i++) { int j = i + len - 1; for(int k = i + 1; k <= j; ++k) { if(str[i] == '[' && str[k] == ']' || str[i] == '(' && str[k] == ')') dp[i][j] = max(dp[i][j], dp[i + 1][k - 1] + dp[k + 1][j] + 2); else dp[i][j] = max(dp[i][j], dp[i][k - 1] + dp[k][j]); } } } printf("%d\n", dp[1] ); } return 0; }
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