【贪心】 Balanced Sequence HDU6299 2018多校第一场

最新推荐文章于 2019-03-18 20:48:45 发布

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最新推荐文章于 2019-03-18 20:48:45 发布

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探讨了如何通过重新排列和连接字符串来最大化平衡子序列长度的问题，介绍了一种有效的算法实现。

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Balanced Sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4423 Accepted Submission(s): 1148

Problem Description

Chiaki has n strings s1,s2,…,sn consisting of '(' and ')'. A string of this type is said to be balanced:

+ if it is the empty string
+ if A and B are balanced, AB is balanced,
+ if A is balanced, (A) is balanced.

Chiaki can reorder the strings and then concatenate them get a new string t. Let f(t) be the length of the longest balanced subsequence (not necessary continuous) of t. Chiaki would like to know the maximum value of f(t) for all possible t.

Input

There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:
The first line contains an integer n (1≤n≤105) -- the number of strings.
Each of the next n lines contains a string si (1≤|si|≤105) consisting of `(' and `)'.
It is guaranteed that the sum of all |si| does not exceeds 5×106.

Output

For each test case, output an integer denoting the answer.

Sample Input

)()(()(

)

)(

Sample Output

Source

2018 Multi-University Training Contest 1

#include <bits/stdc++.h>
using namespace std;

const int mn = 100010;

char ch[mn];
struct node
{
    int l, r;
    int cha;
} a[mn];
bool cmp(const node& a, const node& b)
{
    if (a.cha >= 0 && b.cha < 0)  /// 正贡献左括号，排前面
        return a.cha > b.cha;
    else if (a.cha < 0 && b.cha >= 0)  /// 否则倒置
        return 0;
    else if (a.cha >= 0 && b.cha >= 0)  /// 右括号少的在前
        return a.r < b.r;
    else if (a.cha < 0 && b.cha < 0)
        return a.l > b.l;
}
int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        int n;
        scanf("%d", &n);
        int ans = 0;
        for (int i = 1; i <= n; i++)  // o(5 * 1e6)
        {
            scanf("%s", ch);

            stack<char> s;
            int len = strlen(ch);
            for (int j = 0; j < len; j++)
            {
                if (s.empty())
                {
                    s.push(ch[j]);
                    continue;
                }
                if (ch[j] == '(')
                    s.push(ch[j]);
                else if (ch[j] == ')')
                {
                    if (s.top() == '(')
                    {
                        s.pop();
                        ans += 2;
                    }
                    else
                        s.push(ch[j]);
                }
            }
            a[i].l = 0, a[i].r = 0;
            while (!s.empty())
            {
                if (s.top() == ')')
                    a[i].r++;
                else if(s.top() == '(')
                    a[i].l++;
                s.pop();
            }
            a[i].cha = a[i].l - a[i].r;
        }

        sort(a + 1, a + n + 1, cmp);

        stack<char> s;
        for (int i = 1; i <= n; i++)    // O(5 * 10e6)
        {
            for (int j = 1; j <= a[i].r; j++)
            {
                if (s.empty())
                {
                    s.push(')');
                    continue;
                }
                if (s.top() == '(')
                {
                    s.pop();
                    ans += 2;
                }
                else
                    s.push(')');
            }
            for (int j = 1; j <= a[i].l; j++)
            {
                s.push('(');
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}