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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5083 | Accepted: 2733 |
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
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; char buf[105]; int dp[105][105]; int main() { while(scanf("%s",buf)!=EOF&&buf[0]!=‘e‘) { memset(dp,0,sizeof(dp)); int len=strlen(buf); for(int l=1;l<len;l++)//区间长度 { for(int i=0,j=l;j<len;i++,j++) { if((buf[i]==‘(‘&&buf[j]==‘)‘)||(buf[i]==‘[‘&&buf[j]==‘]‘)) dp[i][j]=dp[i+1][j-1]+2; for(int k=i+1;k<=j-1;k++)//"()[]()" dp[i+1][j-1]=2 而dp[i][j]=6 dp[i][j]=max(dp[i][j],dp[i][k]+dp[k][j]); } } printf("%d\n",dp[0][len-1]); } return 0; }
一种普适的字符串DP顺序
#include<cstdio>
#include<cstring> #include<algorithm> using namespace std;char buf[105]; int dp[105][105]; int main() { while(scanf("%s",buf)&&buf[0]!=‘e‘) { memset(dp,0,sizeof(dp)); int len=strlen(buf); for(int i=0;i<len;i++) { for(int j=i-1;j>=0;j--) { if((buf[j]==‘(‘&&buf[i]==‘)‘)||(buf[j]==‘[‘&&buf[i]==‘]‘)) dp[j][i]=dp[j+1][i-1]+2; for(int x=j+1;x<i;x++) dp[j][i]=max(dp[j][i],dp[j][x]+dp[x][i]); } } printf("%d\n",dp[0][len-1]); } return 0; }
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原文地址：http://www.cnblogs.com/program-ccc/p/5218133.html
