POJ 2955 Brackets 【区间dp】
2016-10-03 18:55
381 查看
Brackets
Submit Status
Description
We 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
Input
The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters
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
Sample Output
Source
Stanford Local 2004
Time Limit: 1000MS | Memory Limit: 65536KB | 64bit IO Format: %lld & %llu |
Description
We 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
Source
Stanford Local 2004
#include<iostream> #include<cstdio> #include<cmath> #include<cstring> #include<climits> #include<string> #include<queue> #include<stack> #include<set> #include<map> #include<algorithm> using namespace std; #define rep(i,j,k)for(i=j;i<k;i++) #define per(i,j,k)for(i=j;i>k;i--) #define MS(x,y)memset(x,y,sizeof(x)) #define max(a,b) a>b?a:b #define min(a,b) a<b?a:b #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 typedef long long LL; const int INF=0x7ffffff; const int M=100+5; char s[M]; int dp[M][M]; int i,j,k,n,m,t; int cheak(char a,char b) { if(a=='('&&b==')') return 1; if(a=='['&&b==']') return 1; return 0; } int main() { while(~scanf("%s",s),s[0]!='e'){ int len=strlen(s); for(i=0;i<len;i++){ dp[i][i]=0; if(cheak(s[i],s[i+1])) dp[i][i+1]=2; else dp[i][i+1]=0; } for(k=3;k<=len;k++) for(i=0;i+k-1<len;i++){ dp[i][i+k-1]=0; if(cheak(s[i],s[i+k-1])) dp[i][i+k-1]=dp[i+1][i+k-2]+2; for(j=i;j<i+k-1;j++) dp[i][i+k-1]=max(dp[i][i+k-1],dp[i][j]+dp[j+1][i+k-1]); } printf("%d\n",dp[0][len-1]); } return 0; }
相关文章推荐
- poj 2955 Brackets(区间DP)
- poj 2955 Brackets 区间dp
- POJ 2955 Brackets(区间DP)
- POJ 2955 Brackets 区间DP
- POJ 2955 Brackets(括号最大匹配,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 入门
- POJ2955——Brackets(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)
- poj 2955 Brackets(区间dp)
- POJ 2955 Brackets(区间DP水题)
- POJ 2955 Brackets 区间DP 入门
- Brackets(POJ - 2955)区间dp
- poj 2955 Brackets(区间DP)
- poj 2955 Brackets 【区间DP】