POJ-2955括号匹配问题（区间DP）-白红宇

POJ-2955括号匹配问题（区间DP）

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发布时间：2019-06-13

本文共 2743 字，大约阅读时间需要 9 分钟。

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4834		Accepted: 2574

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_m where 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

6 4 0 6 Difficulty: （1）.状态：dp[i][j]为i~j的最大括号数。             （2）. 转移：考虑第i个括号，有两种情况：

1.i无效，直接算dp[i + 1][j];                       2.找到和i匹配的右括号k，分两边算并加起来。dp[i][j] = dp[i+1][k-1] + 2 + dp[k + 1][j] 感想：记忆化搜索实质上就是暴力枚举。

#include 
     
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                   using namespace std;const int INF=0x3f3f3f3f;const double eps=1e-10;const double PI=acos(-1.0);#define maxn 1100char a[maxn];int dp[maxn][maxn];int is(char b, char c){ if(b == '(' && c == ')' || b == '[' && c == ']') return 1; else return 0;}int dfs(int st, int ed){ // if(st > ed) return 0; if(st == ed) return 0; if(dp[st][ed] != -1) return dp[st][ed]; int res = dfs(st+1, ed); for(int k = st+1; k <= ed; k++) if(is(a[st],a[k])) { res = max(res,dfs(st+1,k-1) + 2 + dfs(k+1,ed)); flag = 1; } dp[st][ed] = res; return dp[st][ed];}int main(){ while(~scanf("%s", a)) { if(strcmp(a, "end") == 0) break; memset(dp, -1, sizeof dp); int ed = strlen(a)-1; printf("%d\n", dfs(0, ed)); } return 0;}

转载于:https://www.cnblogs.com/ZP-Better/p/5133600.html

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