Brackets
POJ - 2955We 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 [([])]
.
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.
For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.
((())) ()()() ([]]) )[)( ([][][) endSample Output
6 6 4 0 6
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; char ch[105]; int sum[105][105]; int main() { while(~scanf("%s",ch+1)) { if(strcmp(ch+1,"end")==0) break; memset(sum,0,sizeof sum); int len = strlen(ch + 1); for(int l=2;l<=len;l++) //长度 { for(int s=1;s<=len-l+1;s++) //起点 { int e=s+l-1; if((ch[s]=='('&&ch[e]==')')||(ch[s]=='['&&ch[e]==']')) { sum[s][e]=max(sum[s][e],sum[s+1][e-1]+2); } for(int k=s;k<e;k++) //可切割断点k { sum[s][e]=max(sum[s][e],sum[s][k]+sum[k+1][e]); } } } printf("%d\n",sum[1][len]); } return 0; }
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; char ch[105]; int sum[105][105]; int flag[105][105]; int f(int L, int R) { if(R <= L) return 0; if(flag[L][R]) return sum[L][R]; int ans = 0; if(ch[L] == '(' && ch[R] == ')') ans = f(L + 1, R - 1) + 2; if(ch[L] == '[' && ch[R] == ']') ans = f(L + 1, R - 1) + 2; for(int k = L; k < R; k ++) { ans = max(ans, f(L, k) + f(k + 1, R)); } flag[L][R] = 1; return sum[L][R] = ans; } int main() { while(~scanf("%s",ch+1)) { if(strcmp(ch+1,"end")==0) break; memset(sum,0,sizeof sum); memset(flag , 0, sizeof flag); int len = strlen(ch + 1); printf("%d\n", f(1, len)); } return 0; }