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POJ 2955-Brackets

2014-04-19 12:41 411 查看

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Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2611		Accepted: 1349

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 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

Source

Stanford Local 2004

一道很古典的区间DP。。。。水过。。。
dp[ sta] [ed] = { if(两端匹配 ) min(dp[sta+1][ed-1]+2,dp[sta][k]+dp[k+1][ed]) (sta<=k<ed)
else min(dp[sta][k]+dp[k+1][ed]) (sta<=k<ed)
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <map>
using namespace std;
const int maxn = 100+10;
string st;
int n;
int dp[maxn][maxn];
map<char,char> mma;

int dfs(int sta,int ed){
if(sta>=ed) return 0;
if(dp[sta][ed] != -1) return dp[sta][ed];
int ans = 0;
if(mma[st[sta]]==st[ed] ) ans = dfs(sta+1,ed-1)+2;
for(int i = sta; i < ed; i++){
ans = max(ans,dfs(sta,i)+dfs(i+1,ed));
}
return dp[sta][ed] = ans;
}

int main(){
mma['('] = ')';
mma['['] = ']';
while(cin >> st && st != "end"){
memset(dp,-1,sizeof dp);
n = st.size();
cout<<dfs(0,n-1)<<endl;
}
return 0;
}

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