[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。
  1     /*
  2 ━━━━━┒ギリギリ♂ eye！
  3 ┓┏┓┏┓┃キリキリ♂ mind！
  4 ┛┗┛┗┛┃＼○／
  5 ┓┏┓┏┓┃ /
  6 ┛┗┛┗┛┃ノ)
  7 ┓┏┓┏┓┃
  8 ┛┗┛┗┛┃
  9 ┓┏┓┏┓┃
 10 ┛┗┛┗┛┃
 11 ┓┏┓┏┓┃
 12 ┛┗┛┗┛┃
 13 ┓┏┓┏┓┃
 14 ┃┃┃┃┃┃
 15 ┻┻┻┻┻┻
 16 */
 17 #include <algorithm>
 18 #include <iostream>
 19 #include <iomanip>
 20 #include <cstring>
 21 #include <climits>
 22 #include <complex>
 23 #include <fstream>
 24 #include <cassert>
 25 #include <cstdio>
 26 #include <bitset>
 27 #include <vector>
 28 #include <deque>
 29 #include <queue>
 30 #include <stack>
 31 #include <ctime>
 32 #include <set>
 33 #include <map>
 34 #include <cmath>
 35 using namespace std;
 36 #define fr first
 37 #define sc second
 38 #define cl clear
 39 #define BUG puts("here!!!")
 40 #define W(a) while(a--)
 41 #define pb(a) push_back(a)
 42 #define Rint(a) scanf("%d", &a)
 43 #define Rll(a) scanf("%lld", &a)
 44 #define Rs(a) scanf("%s", a)
 45 #define Cin(a) cin >> a
 46 #define FRead() freopen("in", "r", stdin)
 47 #define FWrite() freopen("out", "w", stdout)
 48 #define Rep(i, len) for(int i = 0; i < (len); i++)
 49 #define For(i, a, len) for(int i = (a); i < (len); i++)
 50 #define Cls(a) memset((a), 0, sizeof(a))
 51 #define Clr(a, x) memset((a), (x), sizeof(a))
 52 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 53 #define lrt rt << 1
 54 #define rrt rt << 1 | 1
 55 #define pi 3.14159265359
 56 #define RT return
 57 #define lowbit(x) x & (-x)
 58 #define onenum(x) __builtin_popcount(x)
 59 typedef long long LL;
 60 typedef long double LD;
 61 typedef unsigned long long ULL;
 62 typedef pair<int, int> pii;
 63 typedef pair<string, int> psi;
 64 typedef pair<LL, LL> pll;
 65 typedef map<string, int> msi;
 66 typedef vector<int> vi;
 67 typedef vector<LL> vl;
 68 typedef vector<vl> vvl;
 69 typedef vector<bool> vb;
 70 
 71 const int maxn = 110;
 72 string s, t, f;
 73 bool vis[maxn];
 74 int n, m, k;
 75 int ret;
 76 
 77 int check(int p, char c) {
 78     t.cl(); f.cl();
 79     if(p == 0) {
 80         t.pb(c);
 81         Rep(i, n) t.pb(s[i]);
 82     }
 83     else if(p == n) {
 84         t = s;
 85         t.pb(c);
 86     }
 87     else {
 88         Rep(i, p) t.pb(s[i]);
 89         t.pb(c);
 90         For(i, p, n) t.pb(s[i]);
 91     }
 92     while(1) {
 93         bool exflag = 1;
 94         m = t.length();
 95         f.cl(); Cls(vis);
 96         Rep(i, m-1) {
 97             if(t[i] == t[i+1]) {
 98                 vis[i] = 1;
 99                 vis[i+1] = 1;
100                 exflag = 0;
101             }
102         }
103         Rep(i, m) {
104             if(!vis[i]) {
105                 f.pb(t[i]);
106             }
107         }
108         t = f;
109         if(exflag) break;
110     }
111     return n + 1 - m;
112 }
113 
114 int main() {
115     // FRead();
116     int T;
117     Rint(T);
118     W(T) {
119         cin >> s;
120         n = s.length();
121         ret = 0;
122         For(i, 0, n+1) {
123             ret = max(ret, check(i, 'A'));
124             ret = max(ret, check(i, 'B'));
125             ret = max(ret, check(i, 'C'));
126         }
127         printf("%d
", ret);
128     }
129     RT 0;
130 }