Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 2446 | Accepted: 1258 |
Description
We give the following inductive definition of 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
题意:求最长合法括号序列,
两种写法:
/* *********************************************** Author :rabbit Created Time :2014/3/10 19:03:58 File Name :A.cpp ************************************************ */ #pragma comment(linker, "/STACK:102400000,102400000") #include <stdio.h> #include <iostream> #include <algorithm> #include <sstream> #include <stdlib.h> #include <string.h> #include <limits.h> #include <string> #include <time.h> #include <math.h> #include <queue> #include <stack> #include <set> #include <map> using namespace std; #define INF 0x3f3f3f3f #define eps 1e-8 #define pi acos(-1.0) typedef long long ll; int dp[300][300]; char str[400]; bool match(char a,char b){ if(a==‘(‘&&b==‘)‘)return 1; if(a==‘[‘&&b==‘]‘)return 1; return 0; } int main(){ while(~scanf("%s",str)&&str[0]!=‘e‘){ int len=strlen(str); memset(dp,0,sizeof(dp)); for(int i=0;i+1<len;i++) if(match(str[i],str[i+1]))dp[i][i+1]++; for(int k=2;k<len;k++) for(int i=0;i<len-k;i++){ int j=i+k; if(match(str[i],str[j]))dp[i][j]=dp[i+1][j-1]+1; for(int g=i;g<j;g++) dp[i][j]=max(dp[i][j],dp[i][g]+dp[g+1][j]); } cout<<2*dp[0][len-1]<<endl; } return 0; }
/* *********************************************** Author :rabbit Created Time :2014/3/10 19:03:58 File Name :A.cpp ************************************************ */ #pragma comment(linker, "/STACK:102400000,102400000") #include <stdio.h> #include <iostream> #include <algorithm> #include <sstream> #include <stdlib.h> #include <string.h> #include <limits.h> #include <string> #include <time.h> #include <math.h> #include <queue> #include <stack> #include <set> #include <map> using namespace std; #define INF 0x3f3f3f3f #define eps 1e-8 #define pi acos(-1.0) typedef long long ll; int dp[300][300]; char str[400]; int dfs(int l,int r){ int &res=dp[l][r]; if(res!=-1)return res; if(r-l<2)return res=0; for(int i=l+1;i<r;i++) res=max(res,dfs(l,i)+dfs(i,r)); if((str[l]==‘(‘&&str[r-1]==‘)‘)||(str[l]==‘[‘&&str[r-1]==‘]‘)) res=max(res,dfs(l+1,r-1)+2); return res; } int main() { //freopen("data.in","r",stdin); //freopen("data.out","w",stdout); while(~scanf("%s",str)){ if(strcmp(str,"end")==0)break; int len=strlen(str); memset(dp,-1,sizeof(dp)); int res=dfs(0,len); if(res==-1)res=0; cout<<res<<endl; } return 0; }
原文:http://blog.csdn.net/xianxingwuguan1/article/details/20946345