hihocoder_offer收割编程练习赛55_2

题目链接： https://hihocoder.com/contest/offers55/problem/2

解题思路： 可以发现，这种朋友关系，没什么传递性之类的特征。只能考虑暴力统计，但是20000个字符串之间的比较计算太大，考虑到字符串的长度是10，所有可以查看每个字符串的变化有哪些。

#include <bits/stdc++.h>
using namespace std;

string s[20005];
int a[20005];
int n;
map<string, int> mp;

int judge(string s1, string s2)
{
    int p1=-1, p2=-1;
    int len1 = s1.length();
    int len2 = s2.length();
    if (len1 != len2) return 0;
    for (int i = 0; i < len1; ++i)
    {
        if (s1[i] != s2[i])
        {
            if (p1 == -1)
            {
                p1 = i;
            }
            else if (p2 == -1)
            {
                p2 = i;
            }
            else
            {
                return 0;
            }
        }
    }
    //printf("p1=%d p2=%d\n", p1, p2);
    return s1[p1] == s2[p2] && s1[p2] == s2[p1];
}

int main()
{
    //cout<<judge("aaab", "aaba")<<endl;
    scanf("%d", &n);
    long long ans = 0;
    mp.clear();
    for (int i = 1; i <= n; ++i)
    {
        cin>>s[i];
        if (mp.find(s[i]) == mp.end())
        {
            mp[s[i]] = 1;
        }
        else
        {
            mp[s[i]] += 1;
        }
        //cout<<"insert:"<<s[i]<<endl;
    }
    int sn = s[1].length();
    //printf("sn= %d\n", sn);
    for (auto it = mp.begin(); it != mp.end(); ++it)
    {
        //printf("%s\n", (*it).first.c_str());
        char s[15];
        const char *ts = (*it).first.c_str();
        strcpy(s, ts);
        //printf("s=%s\n", s);
        for (int i = 0; i < sn; ++i)
        {
            for (int j = i+1; j < sn; ++j)
            {
                if (s[i] != s[j])
                {
                    swap(s[i], s[j]);
                    //printf("search=%s\n", s);
                    if (mp.find(string(s)) != mp.end())
                    {
                        ans += mp[string(s)];
                    }
                    swap(s[i], s[j]);
                }
            }
        }
    }
    printf("%lld\n", ans/2);
    return 0;
}