POJ 2955-Brackets（区间DP）

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3340 Accepted: 1716

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 … a_n, 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, …, i_mwhere 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im 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

66406

经典区间DP。

题意：求括号匹配数。

dp[l][r]为区间[l,r]内括号匹配数，dp[l][r]=max(dp[l][r],dp[l+1][k-1]+dp[k+1][r]+2)(s[l]和s[k]匹配)

枚举区间，形式大都一样，3层for循环，看到很多人说区间DP写成记忆化搜索比较容易懂。。orz

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#include <cctype>#include <cmath>#include <cstdlib>#include <vector>#include <queue>#include <set>#include <map>#include <list>#define ll __int64#define pp pair<int,int>using namespace std;const int INF = 0x3f3f3f3f;char s[110];int dp[110][110];void solve(){memset(dp, 0, sizeof(dp));int len = strlen(s) - 1;for (int p = 1; p <= len; p++) {for (int l = 1; l <= len; l++) {int r = l + p - 1;if (r > len) {break;}dp[l][r] = dp[l + 1][r];for (int k = l + 1; k <= r; k++)if ((s[l] == '(' && s[k] == ')') || (s[l] == '[' && s[k] == ']')) {dp[l][r] = max(dp[l][r], dp[l + 1][k - 1] + dp[k + 1][r] + 2);}}}printf("%d\n", dp[1][len]);}int main(){while (~scanf("%s", s + 1) && s[1] != 'e') {s[0] = 2;solve();}return 0;}