Problem - 922D、Robot Vacuum Cleaner - Codeforces

Pushok the dog has been chasing Imp for a few hours already.

Fortunately, Imp knows that Pushok is afraid of a robot vacuum cleaner.While moving, the robot generates a string t consisting of letters ‘s’ and ‘h’, that produces a lot of noise. We define noise of string t as the number of occurrences of string “sh” as a subsequence in it, in other words, the number of such pairs (i, j), that i < j and and .The robot is off at the moment. Imp knows that it has a sequence of strings ti in its memory, and he can arbitrary change their order. When the robot is started, it generates the string t as a concatenation of these strings in the given order. The noise of the resulting string equals the noise of this concatenation.Help Imp to find the maximum noise he can achieve by changing the order of the strings.

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the number of strings in robot’s memory.Next n lines contain the strings t1, t2, …, tn, one per line. It is guaranteed that the strings are non-empty, contain only English letters ‘s’ and ‘h’ and their total length does not exceed 105.

Output

Print a single integer — the maxumum possible noise Imp can achieve by changing the order of the strings.

Examples

input

4
ssh
hs
s
hhhs

output

18

问题解析

这题是说，给你n个字符串，这些字符串全都是由字符’s’和‘h’组成的。让你把这些字符串任意顺序组合起来，使得拼成的字符串含有的’sh‘子序列最多。

关于这题，用到的是一种叫“邻项交换”的方法写的，想法就是这样：两个字符串a和b，如果a+b组成的字符串含有的’sh‘子序列比b+a组成的字符串含有的多，那么最后拼接时把a放在b前是更好的选择。所以这种方法就是让我们写个cmp然后sort一下就行。

cmp内容就是把两个字符串按照不同顺序组合起来，分别计算出他们含有的’sh‘个数，至于’sh‘序列个数，就是每个h字符前面的所有s字符数量加起来。注意一下，如果两个字符串的’sh‘序列数相等，那要返回false（我也不知道为啥，但就是要这样，不然会re）。

AC代码

#include<iostream>
using namespace std;
#include<vector>
#include<algorithm>
#include<math.h>
#include<set>
#include<numeric>
#include<string>
#include<string.h>
#include<iterator>
#include<map>
#include<unordered_map>
#include<stack>
#include<list>
#include<queue>
#include<iomanip>

#define endl '\n'
#define int ll
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PII;
const int N = 2e5 + 50;

int get_sh(string& s)
{
    int cnt = 0, nums = 0, numh = 0;
    for (auto i : s)
    {
        if (i == 's')
        {
            nums++;
        }
        else
        {
            numh++;
            cnt += nums;
        }
    }
    return cnt;
}

bool cmp(string a, string b)
{
    string s1 = a + b, s2 = b + a;
    return get_sh(s1)>get_sh(s2);
}

signed main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int n;
    cin >> n;
    vector<string>v(n);
    for (int i = 0; i < n; i++)
        cin >> v[i];
    sort(v.begin(), v.end(),cmp);
    string s;
    for (auto i : v)s += i;
    cout << get_sh(s) << endl;
    return 0;
}

结尾

邻项交换法也是比较常见的贪心方法了，类似的题目还有算法提高 最小字符串和新国王游戏 - 题目 - Daimayuan Online Judge