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Brackets
| Time Limit: 1000MS |
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Memory Limit: 65536K |
| Total Submissions: 3951 |
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Accepted: 2078 |
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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, andif
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
Source
Stanford Local 2004
题目链接：http://poj.org/problem?id=2955
题目大意：给一个括号序列，问序列中合法的括号最多有多少个，若A合法，则[A]，(A)均合法，若A,B合法则AB也合法
题目分析：和POJ 1141那道经典括号匹配类似，这题更简单一些，想办法把问题转化，既然要求最大的括号匹配数，我们考虑加最少的括号，使得整个序列合法，这样就转变成1141那题，开下脑动类比二分图最大匹配的性质，最大匹配＋最大独立集＝点数，显然要加入最少的点使序列合法，则加的最少的点数即为|最大独立集|，我们要求的是原序列的|最大匹配|，以上纯属yy，下面给出转移方程，和1141一模一样
dp[i][i] = 1;
然后枚举区间长度
1）外围匹配：dp[i][j] = dp[i + 1][j - 1];
2）外围不匹配，枚举分割点：dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]); (i <= k < j)
#include
#include
#include
using namespace std; int const INF = 0x3fffffff; char s[205]; int dp[205][205]; int main() { while(scanf("%s", s) != EOF && strcmp(s, "end") != 0) { int len = strlen(s); memset(dp, 0, sizeof(dp)); for(int i = 0; i < len; i++) dp[i][i] = 1; for(int l = 1; l < len; l++) { for(int i = 0; i < len - l; i++) { int j = i + l; dp[i][j] = INF; if((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']')) dp[i][j] = dp[i + 1][j - 1]; for(int k = i; k < j; k++) dp[i][j] = min(dp[i][j], dp[i][k] + dp[k + 1][j]); } } printf("%d\n", len - dp[0][len - 1]); } }
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