[HIHO1039]字符消除(字符串，枚举，模拟)

题目链接：http://hihocoder.com/problemset/problem/1039

思路：枚举所有字符更新的位置和ABC三种修改方案，之后再模拟消除规则，一步一步去消除。直到无法消除，用原串长度减去当前串长度，更新答案。竟然写了好久。

     /*
 ━━━━━┒ギリギリ♂ eye！
 ┓┏┓┏┓┃キリキリ♂ mind！
 ┛┗┛┗┛┃＼○／
 ┓┏┓┏┓┃ /
 ┛┗┛┗┛┃ノ)
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┛┗┛┗┛┃
 ┓┏┓┏┓┃
 ┃┃┃┃┃┃
 ┻┻┻┻┻┻
 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%lld", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef pair<LL, LL> pll;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 const int maxn = ;
 string s, t, f;
 bool vis[maxn];
 int n, m, k;
 int ret;

 int check(int p, char c) {
     t.cl(); f.cl();
     if(p == ) {
         t.pb(c);
         Rep(i, n) t.pb(s[i]);
     }
     else if(p == n) {
         t = s;
         t.pb(c);
     }
     else {
         Rep(i, p) t.pb(s[i]);
         t.pb(c);
         For(i, p, n) t.pb(s[i]);
     }
     while() {
         bool exflag = ;
         m = t.length();
         f.cl(); Cls(vis);
         Rep(i, m-) {
             if(t[i] == t[i+]) {
                 vis[i] = ;
                 vis[i+] = ;
                 exflag = ;
             }
         }
         Rep(i, m) {
             if(!vis[i]) {
                 f.pb(t[i]);
             }
         }
         t = f;
         if(exflag) break;
     }
     return n +  - m;
 }

 int main() {
     // FRead();
     int T;
     Rint(T);
     W(T) {
         cin >> s;
         n = s.length();
         ret = ;
         For(i, , n+) {
             ret = max(ret, check(i, 'A'));
             ret = max(ret, check(i, 'B'));
             ret = max(ret, check(i, 'C'));
         }
         printf("%d\n", ret);
     }
     RT ;
 }