POJ 2955 区间dp

Brackets

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 2446 Accepted: 1258

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 …a_n, your goal is to find the length of the longest regular brackets sequence that is a subsequence ofs. That is, you wish to find the largest m such that for indicesi₁, i₂, …, i_m where 1 ≤i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 …a_im 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

66406

题意：求最长合法括号序列，

两种写法：

/* ***********************************************Author :rabbitCreated Time :2014/3/10 19:03:58File 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 :rabbitCreated Time :2014/3/10 19:03:58File 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;}