标签：strlen output nbsp find tput sequence [] sse cal
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8017 | Accepted: 4257 |
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
Source
#include<iostream> #include<cstdio> #include<cstring> using namespace std; char s[101]; int dp[101][101]; int main() { while(gets(s)!=NULL) { if(s[0]==‘e‘)break; memset(dp,0,sizeof(dp)); int len=strlen(s); for(int i=1;i<=len;i++) for(int j=0,k=i;k<=len;j++,k++) { if(s[j]==‘(‘&&s[k]==‘)‘||s[j]==‘[‘&&s[k]==‘]‘) dp[j][k]=dp[j+1][k-1]+2; for(int p=j;p<=k;p++) dp[j][k]=max(dp[j][k],dp[j][p]+dp[p+1][k]); } printf("%d\n",dp[0][len-1]); } return 0; }
标签：strlen output nbsp find tput sequence [] sse cal
原文地址：http://www.cnblogs.com/zzyh/p/7157800.html
