poj 2955 Brackets (区间dp基础题)
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
题目大意:
给你一串由 [ ] ( ) 组成的字符串。问最大匹配长度。
区间dp基础题。
dp[i][j]存储i..j字符串的最大匹配长度。预处理出长度为2的区间匹配长度。
dp[i][j]有两种更新方式:如果s[i]与s[j]匹配,那么dp[i][j]=2+dp[i+1][j-1];接着枚举i..j的所有间断点k:dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j])。
区间dp常规操作:
枚举区间长度、枚举左端点、计算右端点、枚举间断点并merge。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn=100; char s[maxn+5]; int dp[maxn+5][maxn+5]; int main() { while(scanf("%s",s+1),s[1]!='e') { int n=strlen(s+1); //预处理出长度为2的区间匹配长度,其余区间全都预置为0 memset(dp,0,sizeof(dp)); for(int i=1;i<n;i++) { if((s[i]=='('&&s[i+1]==')')||(s[i]=='['&&s[i+1]==']')) dp[i][i+1]=2; } //区间DP for(int len=3;len<=n;len++) { for(int left=1;left<=n;left++) { int right=left+len-1; if(right>n) break; //两种状态转移情况都可能发生 if((s[left]=='('&&s[right]==')')||(s[left]=='['&&s[right]==']')) dp[left][right]=2+dp[left+1][right-1]; for(int k=left;k<right;k++) dp[left][right]=max(dp[left][right],dp[left][k]+dp[k+1][right]); //printf("%d %d %d\n",left,right,dp[left][right]); } } printf("%d\n",dp[1][n]); } return 0; }