目录

Brackets (区间DP) 题目题意思路题解

Brackets (区间DP)

题目

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((()))()()()([]]))[)(([][][)endSample Output 6 6 4 0 6

题意

输出能够匹配的最多的括号个数

思路

由小区间的数量推出大区间的括号数量

对于小区间 如果能够首尾匹配那么 dp[i][j]=dp[i+1][j-1]+2 因为这是最直接的。但是这样不是最准确的。对于 ()() 来说 dp[2][3]=0 因为 )( 不匹配。那么用一算出来的 dp[1][4]=0+2=2 是错的，其应该等于 4 .所以这就应该进行扫描子区间。 实际最大值应该是dp[1][4] = max( dp[1][4] , dp[1][2] + dp[3][4] ) = 4

题解

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int d[105][105]; int main() { char s[105]; while (scanf("%s", s + 1), s[1] != 'e') { memset(d, 0, sizeof d); int len = strlen(s + 1); //先枚举小区间，因为后面计算大区间时需要使用小区间 for (int l = 0; l <= len; l++) { for (int i = 1; i + l - 1 <= len; i++) { int j = l + i - 1; //基本的状态转移 if ((s[i] == '(' && s[j] == ')') || (s[i] == '[' && s[j] == ']')) { d[i][j] = d[i + 1][j - 1] + 2; } //为了找到真正最大的值，进行扫描。 for (int k = i; k < j; k++) { d[i][j] = max(d[i][j], d[i][k] + d[k + 1][j]); } } } cout << d[1][len] << endl; } }

转载于:https://www.cnblogs.com/tttfu/p/11291003.html