Ponk Warshall

链接：https://ac.nowcoder.com/acm/contest/7817/H
原字符串与新字符串的字符构成一对，那么如果A-A这种就不需要换，若形如A-B就和B-A交换，但如果没有B-A呢？
系统的归一下类：
A-A 0次
A-B B-A 1次
A-B B-C C-A 2次
A-B B-C C-D D-A 3次
for循环搞定，暴力，没做出来实在反思自己，太愚蠢了！。
代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mk make_pair
const int maxn = 1e6+7;

char s1[maxn];
char s2[maxn];
map<pair<char, char>, int > mp;
char dir[4] = {'A','C','G','T'};

int main()
{
    ll ans = 0;
    scanf("%s", s1);
    scanf("%s", s2);
    int len = strlen(s1);
    for(int i = 0; i < len; i++){
        mp[mk(s1[i], s2[i])]++;
    }

    for(int i = 0; i < 4; i++){
        for(int j = 0; j < 4; j++){
            if(i == j){
                continue;
            }
            int minn = min(mp[mk(dir[i], dir[j])], mp[mk(dir[j], dir[i])]);
            mp[mk(dir[j], dir[i])] -= minn;
            mp[mk(dir[i], dir[j])] -= minn;
            ans += minn;
        }
    }
    for(int i = 0; i < 4; i++) {
        for(int j = 0; j < 4; j++){
            for(int k = 0; k < 4; k++){
                if(i == j || j == k || i == k) continue;
                int x = min(min(mp[mk(dir[i], dir[j])], mp[mk(dir[j], dir[k])]), mp[mk(dir[k], dir[i])]);
                mp[mk(dir[i], dir[j])] -= x;
                mp[mk(dir[j], dir[k])] -= x;
                mp[mk(dir[k], dir[i])] -= x;
                ans += 2*x;
            }
        }
    }

        for(int i = 0; i < 4; i++) {
        for(int j = 0; j < 4; j++){
            for(int k = 0; k < 4; k++){
                for(int r = 0; r < 4; r++) {
                    if(i == j || i == k || i == r || j == k || j == r || k == r) continue;
                    int x = min(min(mp[mk(dir[i], dir[j])], mp[mk(dir[j], dir[k])]), min(mp[mk(dir[k], dir[r])], mp[mk(dir[r], dir[i])] ));
                    mp[mk(dir[i], dir[j])] -= x;
                    mp[mk(dir[j], dir[k])] -= x;
                    mp[mk(dir[k], dir[r])] -= x;
                    mp[mk(dir[r], dir[i])] -= x;
                    ans += 3*x;
                }
            }
        }
    }

    cout << ans << endl;

}