Balanced Sequence（多校，排序规则）

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最新推荐文章于 2022-09-23 21:19:24 发布

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分类专栏：多校联合题目

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给定由'('和')'组成的n个字符串，目标是通过重新排列这些字符串并连接成新的字符串t，找到t中最长的平衡子序列长度。平衡字符串定义包括空字符串、两个平衡字符串的拼接以及在平衡字符串前加一对括号。求所有可能的t中的最大值。题目提供样例输入输出以及问题的关键在于排序规则。

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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

主要是排序规则。

代码：

#include <bits/stdc++.h>

using namespace std;
#define ll long long
const int maxn=1e5+10;
char b[maxn];
ll sum;
struct poin
{
    int l,r;
}G[maxn];
int cmp(poin a,poin b)
{
    int x = min(a.l,b.r);
    int y = min(a.r,b.l);
    return x>y||(x==y&&a.l>b.l);
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        sum=0;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%s",b);
            int le=strlen(b);
            int l=0;
            int r=0;
            for(int j=0;j<le;j++)
            {
                if(b[j]=='(')l++;
                else
                {
                    if(l>0)
                    {
                        l--;
                        sum++;
                    }
                    else r++;
                }
            }
            G[i].l=l;G[i].r=r;
        }
        sort(G+1,G+1+n,cmp);
        int sum_l = 0;
        for(int i=1;i<n;i++)
        {
            //cout<<G[i].l<<" "<<G[i].r<<endl;
            sum+=min(G[i].l,G[i+1].r);
            int o =  G[i+1].r-min(G[i].l,G[i+1].r);
            sum+=min(o,sum_l);
            sum_l-=min(o,sum_l);
            sum_l = sum_l+ G[i].l-min(G[i].l,G[i+1].r);

        }
        printf("%lld\n",sum*2);
    }
    return 0;
}