标签：name cout ons pre mem word sub -o amp
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
区间DP经典例题。因为括号是可以嵌套并列的，所以想到区间DP（才不是因为看到了标签..这个DP数组的含义比较好想到,dp[i][j]表示下标i~j的这一段字符串最多有多少个括号匹配了，按照区间DP的常见套路，第一层枚举区间长度，由于匹配最少需要两个括号，所以len从2开始；第二层枚举左端点，同时也可以根据左端点和长度确定右端点。这里先判断一步，如果s[i]和s[j]匹配了，那么dp[i][j]起码是dp[i+1][j-1]+2（为什么说是起码呢？因为考虑样例的()()()这个序列自然就明白了）
然后找决策点，看看由两个分开的子区间能不能转移到当前并且使得答案更大。最后输出的就是dp[0][len-1]。
#include <iostream> #include <cstring> #include <cmath> #include <cstdio> using namespace std; char s[105]; int dp[105][105]; bool check(int l,int r) { if(s[l]==‘(‘&&s[r]==‘)‘)return true; if(s[l]==‘[‘&&s[r]==‘]‘)return true; return false; } int main() { while(scanf("%s",s)&&s[0]!=‘e‘) { memset(dp,0,sizeof(dp)); int len,i,j,k; for(len=2;len<=strlen(s);len++) { for(i=0;i+len-1<=strlen(s)-1;i++) { int j=i+len-1; if(check(i,j)) { dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2); } for(k=i;k<=j-1;k++) { dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]); } } } cout<<dp[0][strlen(s)-1]<<endl; } return 0; }
标签：name cout ons pre mem word sub -o amp
原文地址：https://www.cnblogs.com/lipoicyclic/p/12516612.html
