思路：容易想到如果存在 "AG" "GA"这种，那一定是先交换这些，可以一次交换解决两个位置，如果不存在前面的情况，我们需要找到类似"AG","GT"这种斜对角能抵消的，得到"AT"然后我们需要马上去找有无"AT","TA"这种情况的。
我们知道"ATCG"只会出现16种字符串长度为2的情况，我们可以暴力枚举情况，直到所有位置都被平行位置，复杂度一定是ok的。
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <set> 5 #include <vector> 6 #include <cmath> 7 #include <map> 8 9 using namespace std; 10 11 #define ll long long 12 #define pb push_back 13 #define fi first 14 #define se second 15 #define lson(x) (x << 1) 16 #define rson(x) (x << 1 | 1) 17 #define mid(l,r) ((l + r) >> 1) 18 19 const int INF = 1e9; 20 map<string, int> mp; 21 22 void fun(string fi, int& ans, int& n){ 23 string a{"aa"}, b{"aa"}, c{"aa"}; 24 string se = "ATCG"; 25 //先匹配"AG" "GA"情况 26 for(int i = 0; i < 4; ++i){ 27 for(int j = 0; j < 4; ++j){ 28 if(i == j) continue; 29 a[0] = se[i]; a[1] = se[j]; 30 b[1] = a[0]; b[0] = a[1]; 31 int cnt = min(mp[a], mp[b]); 32 ans += cnt; 33 mp[a] -= cnt; 34 mp[b] -= cnt; 35 n -= 2 * cnt; 36 } 37 } 38 for(int i = 0; i < 4; ++i){ 39 a[0] = fi[0]; a[1] = se[i]; 40 //"AA"情况 41 if(a[0] == a[1]){ 42 n -= mp[a]; 43 mp[a] = 0; 44 }else{ 45 //"AG" "GT"情况 46 b[0] = a[1]; b[1] = a[0]; 47 int cnt = min(mp[a], mp[b]); 48 ans += cnt; 49 mp[a] -= cnt; 50 mp[b] -= cnt; 51 n -= 2 * cnt; 52 for(int x = 0; x < 4; ++x){ 53 if(se[x] == se[i] || se[x] == fi[0]) continue; 54 b[0] = se[i]; b[1] = se[x]; 55 int cnt = min(mp[a], mp[b]); 56 ans += cnt; 57 mp[a] -= cnt; 58 mp[b] -= cnt; 59 n -= cnt; 60 //试探"AG" "GA"情况 61 c[0] = fi[0]; c[1] = se[x]; 62 mp[c] += cnt; 63 b[0] = c[1]; b[1] = c[0]; 64 cnt = min(mp[b], mp[c]); 65 ans += cnt; 66 mp[b] -= cnt; 67 mp[c] -= cnt; 68 n -= 2 * cnt; 69 } 70 } 71 } 72 // cout << n << endl; 73 } 74 75 void solve(){ 76 77 mp["AA"] = 0; 78 mp["AT"] = 0; 79 mp["AC"] = 0; 80 mp["AG"] = 0; 81 mp["TA"] = 0; 82 mp["TT"] = 0; 83 mp["TC"] = 0; 84 mp["TG"] = 0; 85 mp["CA"] = 0; 86 mp["CT"] = 0; 87 mp["CC"] = 0; 88 mp["CG"] = 0; 89 mp["GA"] = 0; 90 mp["GT"] = 0; 91 mp["GC"] = 0; 92 mp["GG"] = 0; 93 string up, down; 94 cin >> up >> down; 95 int n = up.length(); 96 //cout << "length = " << n << endl; 97 int ans = 0; 98 string tmp{"aa"}; 99 for(int i = 0; i < n; ++i){ 100 tmp[0] = up[i]; 101 tmp[1] = down[i]; 102 ++mp[tmp]; 103 } 104 while(n){ 105 fun("A", ans, n); 106 fun("T", ans, n); 107 fun("C", ans, n); 108 fun("G", ans, n); 109 } 110 111 cout << ans << endl; 112 } 113 114 int main(){ 115 116 ios::sync_with_stdio(false); 117 cin.tie(0); cout.tie(0); 118 // freopen("C:\Users\admin\Desktop\input.txt", "w", stdin); 119 // freopen("C:\Users\admin\Desktop\output.txt", "r", stdout); 120 solve(); 121 122 return 0; 123 }