poj -2955 Brackets
http://poj.org/problem?id=2955
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 2707 | Accepted: 1403 |
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 msuch 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<stdio.h> #include<string.h> int dp[100][100]; int max(int x,int y) { if(x>y) return x; else return y; } bool match(char x,char y) { if(x=='['&&y==']') return true; else if(x=='('&&y==')') return true; else return false; } int main() { int len,i,j,k,g; char str[100]; while(gets(str)) { if(str[0]=='e') break; memset(dp,0,sizeof(dp)); len=strlen(str); for(i=0;i<len;i++) { dp[i][i]=0; if(match(str[i],str[i+1])) dp[i][i+1]=2;//初始值。 } for(k=0;k<len;k++) for(i=0;i<len-k;i++) { j=i+k; if(match(str[i],str[j])) dp[i][j]=dp[i+1][j-1]+2; for(g=0;g<k;g++) dp[i][j]=max(dp[i][j],dp[i][i+g]+dp[i+g+1][j]); } printf("%d\n",dp[0][len-1]); } return 0; }
别人的代码,可供参考。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn=1002; char s[maxn]; int dp[maxn][maxn]; int main() { //freopen("//media/学习/ACM/input.txt","r",stdin); while(scanf("%s",s),s[0]!='e') { int i,j,k,n=strlen(s); for(i=0;i<n;i++) for(j=0;j<n;j++) dp[i][j]=0; for(i=n-1;i>=0;i--) { for(j=i+1;j<=n-1;j++) { dp[i][j]=max(dp[i+1][j],dp[i][j-1]); for(k=i+1;k<=j;k++) { if((s[i]=='('&&s[k]==')')||(s[i]=='['&&s[k]==']')) dp[i][k]=max(dp[i][k],dp[i+1][k-1]+2); dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]); } // cout<<i<<" "<<j<<" "<<dp[i][j]<<endl; } } cout<<dp[0][n-1]<<endl; } return 0; }