多校1——Balanced Sequence（贪心）

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

2

1

)()(()(

2

)

)(

Sample Output

4

2

Source

2018 Multi-University Training Contest 1

Recommend

liuyiding   |   We have carefully selected several similar problems for you:  6308 6307 6306 6305 6304 

这道题的题意就是，括号串可以排序，然后将其连起来，问合法的最长是多少。

这道题当时比赛时就是想的非常复杂，其实觉得自己的思路要是认真的揣摩一下就好了。

所有的括号预处理一下就可以知道，剩下的括号的样子就是“（（（”或者“）））”或者“）））（（（（”

那么将其再细分，就可以演变成几种情况：

①左括号>右括号

②左括号<=右括号

那么一眼就可以看出，左括号越小的，越应该放在最前边，因为左边的括号肯定就浪费了，没法搭配了。

当时比赛时我就想到了这一层。

然后正解就是：

①：左括号<右括号，按照左括号从小到大排

②：左括号>=右括号，按照左括号从大到小排。

这种贪心方法以后遇到类似的，也不妨大胆尝试一番。

代码：

#include
using namespace std;
const int maxn=100010;

struct node
{
    int l,r;
}a[maxn];
int cmp(node a,node b)
{
    if(a.l<=a.r&&b.l>b.r) //)少(多 > )多(少
        return true;///不变
    if(a.l>a.r&&b.l<=b.r) //)多(少 < )少(多
        return false;///让a，b交换
    if(a.l<=a.r&&b.l<=b.r)//)少(多   )少(多
        return a.l=a.r &&b.l>b.r)
        return a.r>b.r;       //(多的优先
}

int main()
{
    int t;
    char H[maxn];
    char h[maxn];
    int n;
    scanf("%d",&t);
    while(t--)
    {
        int ans=0;
        scanf("%d",&n);
        int top;
        for(int i=1;i<=n;i++)
        {
            top=0;
            scanf("%s",H);
            int kaka=strlen(H);
            for(int j=0;j0)
                {
                    ans++;
                    --top;
                }
                else
                {
                    h[++top]=H[j];
                }
            }
                a[i].l=0;
                a[i].r=0;
                for(int j=1;j<=top;j++)
                {
                    if(h[j]==')')
                        a[i].l++;
                    else
                        a[i].r++;
                }
            }

            sort(a+1,a+n+1,cmp);
             int num=0;
            for(int i=2;i<=n;i++)
            {
                num+=a[i-1].r;
                if(a[i].l>num)
                {
                    ans+=num;
                    num=0;
                }
                else
                {
                    ans+=a[i].l;
                    num-=a[i].l;
                }
            }
            printf("%d\n",ans*2);

        }

}