poj 2955 Brackets(区间DP)
2017-08-01 08:54
363 查看
Brackets
DescriptionWe 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 … 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
思路:区间DP
如果((s[i]==’(‘&&s[j]==’)’)||(s[i]==’[‘&&s[j]==’]’)), 很明显这时侯
dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2);
其他情况下状态转移方程:dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
代码:
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; char s[110]; int dp[110][110]; int judge(char a,char b) { if((a=='('&&b==')')||(a=='['&&b==']')) return 1; return 0; } int main() { while(~scanf("%s",s+1)) { memset(dp,0,sizeof(dp)); if(!strcmp(s+1,"end")) break; int len=strlen(s+1); int ans=0; for(int l=1;l<len;++l) for(int i=1;i+l<=len;++i) { int j=i+l; if(judge(s[i],s[j])) dp[i][j]=max(dp[i][j],dp[i+1][j-1]+2); for(int k=i;k<j;++k) dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]); ans=max(ans,dp[i][j]); } printf("%d\n",ans); } return 0; }
相关文章推荐
- POJ 2955 Brackets (区间dp 括号匹配)
- POJ 2955 - Brackets(区间DP)
- poj 2955 Brackets(区间DP)
- POJ 2955 Brackets 【区间DP】
- poj 2955 Brackets 括号匹配 区间dp
- poj 2955 Brackets(区间DP)
- POJ 2955-Brackets(区间DP)
- POJ 2955 Brackets(区间dp)
- POJ 2955 Brackets (区间DP,常规)
- poj 2955 Brackets(区间DP)
- POJ 2955-Brackets(括号匹配-区间DP)
- poj 2955 Brackets 区间DP
- poj 2955 Brackets(区间dp)
- poj 2955 Brackets(区间DP求最长匹配子串)
- poj 2955 Brackets(区间DP)
- POJ 2955-Brackets(区间DP)
- HOJ 1936&POJ 2955 Brackets(区间DP)
- POJ 2955 Brackets 区间DP 入门题
- POJ 2955 Brackets & POJ 1505 Copying Books & POJ 1651 Multiplication Puzzle(初级区间DP)
- poj 2955 Brackets(括号匹配,区间DP)