| 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