| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 14462 | Accepted: 7595 |
Description
We give the following inductive definition of 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：
给你由(,),[,]组成的字符串，从中取出一子串，问最大的括号匹配数？
分析：
求解DP的问题要从三个方面出发：子问题+最优解+极限状态。
设dp[i][j]表示字符串str[i]~str[j]的最大匹配数。
子问题：
（1）外层括号可以包含内层括号，所以当str[i],str[j]匹配时，dp[i][j] = dp[i+1][j-1] + 2;
（2）dp[i][j] 可以被分为很多不同的段的解的和值，即dp[i][j] = dp[i][k]+dp[k+1][j];
最优解：
dp[i][j] 为上面所有情况的最大值。
极限状态：
长度为1时：dp[i][i] = 0; 然后从长度为2~n开始递推出所有解。
#include<iostream> #include<string.h> #include<string> #include<math.h> using namespace std; string s; int dp[105][105]; int main() { while(cin>>s) { if(s=="end") break; memset(dp,0,sizeof(dp)); int n=s.length(); for(int len=2;len<=n;len++)//枚举区间长度 { for(int i=0;i+len<=n;i++)//枚举区间左端点 { int j=i+len-1;//区间右端点 if(j>n)//防止越界 break; if(s[i]==‘(‘&&s[j]==‘)‘||s[i]==‘[‘&&s[j]==‘]‘)//如果第i个和第j个匹配 dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2); for(int k=i;k<j;k++)//如果第 i 个和第 j 个不匹配，枚举中间分割点k dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]); } } cout<<dp[0][n-1]<<endl; } }
原文：https://www.cnblogs.com/-citywall123/p/10896480.html