NAIPC 2017

Yin and Yang Stones

• 75.39%
• 1000ms
• 262144K

A mysterious circular arrangement of black stones and white stones has appeared. Ming has been tasked with balancing the stones so that only one black and one white stone remain.

Ming has two operations for balancing the stones:

1. Take some consecutive sequence of stones where there is exactly one more black stone than a white stone and replace the stones with a single black stone

2. Take some consecutive sequence of stones where there is exactly one more white stone than black stone and replace the stones with a single white stone

Given a circular arrangement, determine if it is possible for Ming to balance the stones.

Input

Each input will consist of a single test case. Note that your program may be run multiple times on different inputs. The input will consist of a single string sss (1≤∣s∣≤105)(1 \le |s| \le 10^5)(1≤∣s∣≤105), with only the characters capital ‘BBB’ and ‘WWW’. The stones are arranged in a circle, so the first stone and the last stone are adjacent.

Output

Output 111 if it is possible for Ming to balance the stones with his rules. Otherwise, output 000.

樣例輸入1

WWBWBB

樣例輸出1

1

樣例輸入2

WWWWBBW

樣例輸出2

0

樣例輸入3

WBBBBBWWBW

樣例輸出3

0

題目來源

The North American Invitational Programming Contest 2017

思路：W與B相同輸出1，否則輸出0。只有這樣才能保持黑白平衡。

Pieces of Parentheses

• 22.03%
• 1000ms
• 262144K

You are teaching a class in programming, and you want to cover balanced parentheses. You’ve got a great visual aid, a sign with a very long, balanced string of parentheses. But, alas, somehow, your visual aid has been broken into pieces, and some pieces may be missing! You’ve got to try to put it back together as best you can. Given the string of parentheses on each piece, what is the longest balanced string you can form by concatenating some of them in some order? Each piece may be used at most once, and the pieces cannot be reversed.

A balanced string of parentheses is defined as:

1. The empty string

2. ABABAB where AAA and BBB are both balanced strings of parentheses

3. (A)(A)(A) where AAA is a balanced string of parentheses

Input

Each input will consist of a single test case. Note that your program may be run multiple times on different inputs. The first line of input will contain a single integer n(1≤n≤300)n (1 \le n \le 300)n(1≤n≤300), which is the number of pieces.

Each of the next nnn lines will hold a single string s(1≤∣s∣≤300)s (1 \le |s| \le 300)s(1≤∣s∣≤300), which consists only of the characters ’(((’ and ’)))’. This describes one of the pieces.

Output

Output a single integer, which is the length of the longest string of balanced parentheses you can form from the pieces. Note that the empty string is technically a balanced string of parentheses, so it is always possible to form a string of length at least 000 (although the empty string is not a very effective visual aid!).

樣例輸入1

3
())
((()
)()

樣例輸出1

10

樣例輸入2

5
)))))
)
((
))((
(

樣例輸出2

2

題目來源

The North American Invitational Programming Contest 2017

代碼：

  1 #include <iostream>
  2 #include <cstring>
  3 #include <queue>
  4 using namespace std;
  5 template <class T, class C>
  6 using heap = priority_queue<T, vector<T>, C>;
  7 void abc(string s, int &a, int &b, int &c)
  8 {
  9     a = 0,
 10     b = 0,
 11     c = s.length();
 12     for (int i = 0; i < s.length(); i++)
 13     {
 14         switch (s[i])
 15         {
 16         case ‘(‘:
 17             a++;
 18             break;
 19         case ‘)‘:
 20             if (a > 0)
 21             {
 22                 a--;
 23             }
 24             else
 25             {
 26                 b++;
 27             }
 28         }
 29     }
 30 }
 31 struct triple
 32 {
 33     int a,
 34         b,
 35         c;
 36 };
 37 bool operator>(const triple &A, const triple &B)
 38 {
 39     if (A.b ^ B.b)
 40     {
 41         return A.b > B.b;
 42     }
 43     if (A.a ^ B.a)
 44     {
 45         return A.a < B.a;
 46     }
 47     return A.c < B.c;
 48 }
 49 bool operator<(const triple &A, const triple &B)
 50 {
 51     if (A.a ^ B.a)
 52     {
 53         return A.a > B.a;
 54     }
 55     if (A.b ^ B.b)
 56     {
 57         return A.b < B.b;
 58     }
 59     return A.c < B.c;
 60 }
 61 int main()
 62 {
 63     int n{0};
 64     cin >> n;
 65     int A[90001], B[90001];
 66     memset(A, 0xf0, sizeof(A));
 67     memset(B, 0xf0, sizeof(B));
 68     A[0] = 0;
 69     B[0] = 0;
 70     heap<triple, greater<triple>> I;
 71     heap<triple, less<triple>> D;
 72     for (int i = 1; i <= n; i++)
 73     {
 74         string s;
 75         cin >> s;
 76         int a{0}, b{0}, c{0};
 77         abc(s, a, b, c);
 78         if (a >= b)
 79         {
 80             I.push({a, b, c});
 81         }
 82         else
 83         {
 84             D.push({a, b, c});
 85         }
 86     }
 87     while (I.size())
 88     {
 89         const int a = I.top().a,
 90                   b = I.top().b,
 91                   c = I.top().c;
 92         for (int x = 90000; x >= max(b, a - b); x--)
 93         {
 94             A[x] = max(A[x], A[x - a + b] + c);
 95         }
 96         I.pop();
 97     }
 98     while (D.size())
 99     {
100         const int a = D.top().a,
101                   b = D.top().b,
102                   c = D.top().c;
103         for (int x = 90000; x >= max(a, b - a); x--)
104         {
105             B[x] = max(B[x], B[x - b + a] + c);
106         }
107         D.pop();
108     }
109     int reponse{0};
110     for (int x = 0; x <= 90000; x++)
111     {
112         reponse = max(reponse, A[x] + B[x]);
113     }
114     cout << reponse << endl;
115     return 0;
116 }

參考博客：http://www.cnblogs.com/JebediahKerman/p/9742462.html

The North American Invitational Programming Contest 2017 題目