CodeForces - 593A - O - 2Char - 暴力枚举

Posted 2021-05-21 Chivas_/Regal

题目：

思路：
今天搞这个1200分的题，有点搞心态了
一个1200的题让我写出了1400的感受
（可能是我太菜了吧）

其实就枚举就行了
首先内部不同字符个数 > 2的都不能存
然后枚举要用哪两个字符
然后看这个字符串是不是只有枚举出来的两个字符之一或都有
至于查看这个字符串有哪些字符的方法：只需要把字符串排个序，内部存的字符就暴露在头和尾啦

代码：

/*
           ________   _                                              ________                              _
          /  ______| | |                                            |   __   |                            | |
         /  /        | |                                            |  |__|  |                            | |
         |  |        | |___    _   _   _   ___  _   _____           |     ___|   ______   _____   ___  _  | |
         |  |        |  __ \\  |_| | | | | |  _\\| | | ____|          |  |\\  \\    |  __  | |  _  | |  _\\| | | |
         |  |        | |  \\ |  _  | | | | | | \\  | | \\___           |  | \\  \\   | |_/ _| | |_| | | | \\  | | |
         \\  \\______  | |  | | | | \\ |_| / | |_/  |  ___/ |          |  |  \\  \\  |    /_   \\__  | | |_/  | | |
Author :  \\________| |_|  |_| |_|  \\___/  |___/|_| |_____| _________|__|   \\__\\ |______|     | | |___/|_| |_|
                                                                                         ____| |
                                                                                         \\_____/
*/
#include <unordered_map>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <cstdio>
#include <stack>
#include <queue>
#include <cmath>
#include <utility>
#include <map>
#include <set>

#define G 10.0
#define LNF 1e18
#define EPS 1e-6
#define PI acos(-1.0)
#define INF 0x7FFFFFFF

#define ll long long
#define ull unsigned long long

#define LOWBIT(x) ((x) & (-x))
#define LOWBD(a, x) lower_bound(a.begin(), a.end(), x) - a.begin()
#define UPPBD(a, x) upper_bound(a.begin(), a.end(), x) - a.begin()
#define TEST(a) cout << "---------" << a << "---------" << '\\n'

#define CHIVAS int main()
#define _REGAL exit(0)

#define SP system("pause")
#define IOS ios::sync_with_stdio(false)
#define map unordered_map

#define PB(x) push_back(x)
#define ALL(a) ((a).begin(),(a).end())
#define MEM(a, b) memset(a, b, sizeof(a))
#define EACH_CASE(cass) for (cin >> cass; cass; cass--)

#define LS l, mid, rt << 1
#define RS mid + 1, r, rt << 1 | 1
#define GETMID (l + r) >> 1

using namespace std;

template<typename T> inline T MAX(T a, T b){return a > b? a : b;}
template<typename T> inline T MIN(T a, T b){return a > b? b : a;}
template<typename T> inline void SWAP(T &a, T &b){T tp = a; a = b; b = tp;}
template<typename T> inline T GCD(T a, T b){return b > 0? GCD(b, a % b) : a;}
template<typename T> inline void ADD_TO_VEC_int(T &n, vector<T> &vec){vec.clear(); cin >> n; for(int i = 0; i < n; i ++){T x; cin >> x, vec.PB(x);}}
template<typename T> inline pair<T, T> MaxInVector_ll(vector<T> vec){T MaxVal = -LNF, MaxId = 0;for(int i = 0; i < (int)vec.size(); i ++) if(MaxVal < vec[i]) MaxVal = vec[i], MaxId = i; return {MaxVal, MaxId};}
template<typename T> inline pair<T, T> MinInVector_ll(vector<T> vec){T MinVal = LNF, MinId = 0;for(int i = 0; i < (int)vec.size(); i ++) if(MinVal > vec[i]) MinVal = vec[i], MinId = i; return {MinVal, MinId};}
template<typename T> inline pair<T, T> MaxInVector_int(vector<T> vec){T MaxVal = -INF, MaxId = 0;for(int i = 0; i < (int)vec.size(); i ++) if(MaxVal < vec[i]) MaxVal = vec[i], MaxId = i; return {MaxVal, MaxId};}
template<typename T> inline pair<T, T> MinInVector_int(vector<T> vec){T MinVal = INF, MinId = 0;for(int i = 0; i < (int)vec.size(); i ++) if(MinVal > vec[i]) MinVal = vec[i], MinId = i; return {MinVal, MinId};}
template<typename T> inline pair<map<T, T>, vector<T> > DIV(T n){T nn = n;map<T, T> cnt;vector<T> div;for(ll i = 2; i * i <= nn; i ++){while(n % i == 0){if(!cnt[i]) div.push_back(i);cnt[i] ++;n /= i;}}if(n != 1){if(!cnt[n]) div.push_back(n);cnt[n] ++;n /= n;}return {cnt, div};}


inline void solve(){
    int n; cin >> n;
    vector<string> vec;
    string s;
    for(int i = 0; i < n; i ++){
        cin >> s;
        sort(s.begin(), s.end());

        map<char, int> mp;
        int cnt = 0;
        for(int j = 0; j < s.size(); j ++){
            if(!mp[s[j]]) mp[s[j]] = 1, cnt ++;
        }if(cnt <= 2) vec.push_back(s);
    }

    int res = 0;//维护最大值
    for(char i = 'a'; i <= 'z'; i ++){
        for(char j = i + 1; j <= 'z'; j ++){
            int cur = 0;//当前答案
            for(int k = 0; k < vec.size(); k ++){
                if(vec[k][0] == vec[k][vec[k].size() - 1]){//头和尾一样的话整个字符串只有一个字符，看这个字符是不是枚举的其中之一
                    if(vec[k][0] == i || vec[k][0] == j) 
                        cur += vec[k].size();
                }else{//如果不一样说明有两个不同的字符，j比i大，所以头要等于i尾要等于j
                    if(vec[k][0] == i && vec[k][vec[k].size() - 1] == j)
                        cur += vec[k].size();
                }
            }
            res = MAX(res, cur);
        }
    }

    cout << res <<endl;
}

CHIVAS{
    solve();
    _REGAL;
}