身体是革命的本钱，这道题是关于运动的。
Alice和Bob在进行乒乓球比赛，比赛一共打了 n 个球，对于每一球，如果Alice赢了，那么裁判员会在计分板上记下'A'，如果Bob赢了则会记下'B'。
时间转眼间到了2050年，计分板上某些信息因为时间流逝丢失了，但我们想要复现当年的激烈局面。
丢失的位置用'?'表示，我们想知道，计分板上对应的乒乓球球赛，最多进行了多少局（最后一局可以没打完，但是如果没打完的话就不计入答案）？
在一局比赛中，先得11分的一方为胜方，10平后，先多得2分的一方为胜方。
第一行一个整数 T (1≤T≤51) 表示数据组数。
接下来 T 组数据，每行一个字符串表示计分板上记录的信息，计分板上只包含'A'，'B'，'?'这些字符，计分板长度 n≤10000。
\(dp_{i,j,k}\) 表示当前弄到第 \(i\) 个位置有 \(j:k\) 最大的局数。
// powered by c++11
// by Isaunoya
#pragma GCC diagnostic error "-std=c++11"
#include <bits/stdc++.h>
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
using namespace std;
using db = double;
using ll = long long;
using uint = unsigned int;
using ull = unsigned long long;
using pii = pair<int, int>;
#define fir first
#define sec second
template <class T>
void cmax(T& x, const T& y) {
if (x < y) x = y;
}
template <class T>
void cmin(T& x, const T& y) {
if (x > y) x = y;
}
#define all(v) v.begin(), v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
template <class T>
void sort(vector<T>& v) {
sort(all(v));
}
template <class T>
void reverse(vector<T>& v) {
reverse(all(v));
}
template <class T>
void unique(vector<T>& v) {
sort(all(v)), v.erase(unique(all(v)), v.end());
}
void reverse(string& s) { reverse(s.begin(), s.end()); }
const int io_size = 1 << 23 | 233;
const int io_limit = 1 << 22;
struct io_in {
char ch;
#ifndef __WIN64
char getchar() {
static char buf[io_size], *p1 = buf, *p2 = buf;
return (p1 == p2) && (p2 = (p1 = buf) + fread(buf, 1, io_size, stdin), p1 == p2) ? EOF : *p1++;
}
#endif
io_in& operator>>(char& c) {
for (c = getchar(); isspace(c); c = getchar());
return *this;
}
io_in& operator>>(string& s) {
for (s.clear(); isspace(ch = getchar());)
;
if (!~ch) return *this;
for (s = ch; !isspace(ch = getchar()) && ~ch; s += ch)
;
return *this;
}
io_in& operator>>(char* str) {
char* cur = str;
while (*cur) *cur++ = 0;
for (cur = str; isspace(ch = getchar());)
;
if (!~ch) return *this;
for (*cur = ch; !isspace(ch = getchar()) && ~ch; *++cur = ch)
;
return *++cur = 0, *this;
}
template <class T>
void read(T& x) {
bool f = 0;
while ((ch = getchar()) < 48 && ~ch) f ^= (ch == 45);
x = ~ch ? (ch ^ 48) : 0;
while ((ch = getchar()) > 47) x = x * 10 + (ch ^ 48);
x = f ? -x : x;
}
io_in& operator>>(int& x) { return read(x), *this; }
io_in& operator>>(ll& x) { return read(x), *this; }
io_in& operator>>(uint& x) { return read(x), *this; }
io_in& operator>>(ull& x) { return read(x), *this; }
io_in& operator>>(db& x) {
read(x);
bool f = x < 0;
x = f ? -x : x;
if (ch ^ '.') return *this;
double d = 0.1;
while ((ch = getchar()) > 47) x += d * (ch ^ 48), d *= .1;
return x = f ? -x : x, *this;
}
} in;
struct io_out {
char buf[io_size], *s = buf;
int pw[233], st[233];
io_out() {
set(7);
rep(i, pw[0] = 1, 9) pw[i] = pw[i - 1] * 10;
}
~io_out() { flush(); }
void io_chk() {
if (s - buf > io_limit) flush();
}
void flush() { fwrite(buf, 1, s - buf, stdout), fflush(stdout), s = buf; }
io_out& operator<<(char c) { return *s++ = c, *this; }
io_out& operator<<(string str) {
for (char c : str) *s++ = c;
return io_chk(), *this;
}
io_out& operator<<(char* str) {
char* cur = str;
while (*cur) *s++ = *cur++;
return io_chk(), *this;
}
template <class T>
void write(T x) {
if (x < 0) *s++ = '-', x = -x;
do {
st[++st[0]] = x % 10, x /= 10;
} while (x);
while (st[0]) *s++ = st[st[0]--] ^ 48;
}
io_out& operator<<(int x) { return write(x), io_chk(), *this; }
io_out& operator<<(ll x) { return write(x), io_chk(), *this; }
io_out& operator<<(uint x) { return write(x), io_chk(), *this; }
io_out& operator<<(ull x) { return write(x), io_chk(), *this; }
int len, lft, rig;
void set(int _length) { len = _length; }
io_out& operator<<(db x) {
bool f = x < 0;
x = f ? -x : x, lft = x, rig = 1. * (x - lft) * pw[len];
return write(f ? -lft : lft), *s++ = '.', write(rig), io_chk(), *this;
}
} out;
#define int long long
template <int sz, int mod>
struct math_t {
math_t() {
fac.resize(sz + 1), ifac.resize(sz + 1);
rep(i, fac[0] = 1, sz) fac[i] = fac[i - 1] * i % mod;
ifac[sz] = inv(fac[sz]);
Rep(i, sz - 1, 0) ifac[i] = ifac[i + 1] * (i + 1) % mod;
}
vector<int> fac, ifac;
int qpow(int x, int y) {
int ans = 1;
for (; y; y >>= 1, x = x * x % mod)
if (y & 1) ans = ans * x % mod;
return ans;
}
int inv(int x) { return qpow(x, mod - 2); }
int C(int n, int m) {
if (n < 0 || m < 0 || n < m) return 0;
return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
};
int gcd(int x, int y) { return !y ? x : gcd(y, x % y); }
int lcm(int x, int y) { return x * y / gcd(x, y); }
const int maxn = 1e4 + 41;
int dp[maxn][12][12];
char s[maxn];
struct node {
node(int a = 0 , int b = 0 , int c = 0) {
x = a , y = b , z = c;
}
int x , y , z;
};
node get(int x , int y) {
if(x == 11 || y == 11) {
return node(0 , 0 , 1);
}
if(x == 10 && y == 10) {
return node(9 , 9 , 0);
}
else {
return node(x , y , 0);
}
}
signed main() {
// code begin.
int _ ;
in >> _;
while(_ --) {
in >> s;
memset(dp, -1, sizeof(dp));
char * cur = s;
int now = 0;
dp[0][0][0] = 0;
while(* cur) {
++ now;
char c = (* cur ++);
rep(i , 0 , 10) {
rep(j , 0 , 10) {
if(~ dp[now - 1][i][j]) {
if(c == 'A' || c == '?') {
auto it = get(i + 1 , j);
dp[now][it.x][it.y] = max(dp[now][it.x][it.y], dp[now - 1][i][j] + it.z);
}
if(c == 'B' || c == '?') {
auto it = get(i , j + 1);
dp[now][it.x][it.y] = max(dp[now][it.x][it.y], dp[now - 1][i][j] + it.z);
}
}
}
}
}
int ans = 0 ;
rep(i , 0 , 10) rep(j , 0 , 10) cmax(ans, dp[now][i][j]);
out << ans << '\n';
}
return 0;
// code end.
}
来源：https://www.cnblogs.com/Isaunoya/p/12591623.html