Codeforces Round #329 (Div. 2) A. 2Char ---- 思维+STL

版权声明：本文为博主原创文章，转载请预先通知博主(〃'▽'〃)。 https://blog.csdn.net/m0_37624640/article/details/82867947

题目传送门

做法：一次性用两个不同的字符去枚举所有串即可，找到最大的

AC代码：跟超霸学的一手好写法( ﾟ∀ﾟ)    

#include<bits/stdc++.h>
#define IO          ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define pb(x)       push_back(x)
#define sz(x)       (int)(x).size()
#define sc(x)       scanf("%d",&x)
#define pr(x)       printf("%d\n",x)
#define abs(x)      ((x)<0 ? -(x) : x)
#define all(x)      x.begin(),x.end()
#define mk(x,y)     make_pair(x,y)
#define debug       printf("!!!!!!\n")
#define fin         freopen("in.txt","r",stdin)
#define fout        freopen("out.txt","w",stdout)
using namespace std;
typedef long long ll;
const int mod = 1e9+7;
const double PI = 4*atan(1.0);
const int maxm = 1e6;
const int maxn = 1e6+5;
const int INF = 0x3f3f3f3f;
const ll LINF = 1ll<<62;
char s[105][1005];
map<char,int> m[105];
int main()
{
    //fin;
    IO;
    int n;
    cin>>n;
    for(int i=0;i<n;i++) cin>>s[i];
    for(int i=0;i<n;i++)
    {
        for(int j=0;s[i][j];j++)
            m[i][s[i][j]]++;
    }
    int ans = 0;
    for(char i = 'a';i<='z';i++) //一次性枚举两个字符
    {
        for(char j=i+1;j<='z';j++) //一次性枚举两个字符
        {
            int tmp = 0;
            for(int k=0;k<n;k++){ //n个字符串
                int nsize = m[k].size();
                if(nsize >=3 ) continue;
                if(nsize == 1){
                    if(m[k].count(i)) tmp+=m[k][i];
                    if(m[k].count(j)) tmp+=m[k][j];
                }
                if(nsize == 2){
                    if(m[k].count(i) && m[k].count(j)) tmp = tmp+m[k][i]+m[k][j];
                }
            }
            ans = max(tmp,ans);
        }
    }
    cout<<ans<<endl;
    return 0;
}