牛客小白月赛116-Poi 的消消乐 · znzryb
## 思路讲解
首先，拿到题目，发现为01串，这个时候就要想到分块思想。
如何选用乘法压缩石块那？更优解。让B作为首位。但这种情况必须后面有A（即 idx≥3）。
最后代码实现就是这样
```cpp
if(idx==1){
cout<<1<<"\n";
}else if(idx==2){
cout<<1+min(SZ(g[st[2]-'A']),3)<<"\n";
}else{
cout<<2<<"\n";
}
```
## AC代码
::: 源代码
```cpp
// Problem: Poi 的消消乐
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/109080/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// by znzryb
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define all(vec) vec.begin(),vec.end()
#define CLR(i,a) memset(i,a,sizeof(i))
#define fi first
#define se second
#define pb push_back
#define SZ(a) ((int) a.size())
#define FOR(i, a, b) for (int i = (a); i <= (b); ++i)
#define ROF(i, a, b) for (int i = (a); i >= (b); --i)
#define getFro(vec) (vec.empty()?0:vec.front())
#define getBac(vec) (vec.empty()?0:vec.front())
#define debug(var) cerr << #var <<":"<<var<<"\n";
#define DEBUG(variable) \
do { \
std::cerr << #variable << ":"; \
for (const auto& elem : variable) { \
std::cerr << elem << " "; \
} \
std::cerr << "\n"; \
} while (0)
#define uniVec(var) \
do { \
sort(var.begin(),var.end());\
var.resize(unique(var.begin(),var.end())-var.begin());\
} while (0)
#define debugN(var,N) \
do{ \
std::cerr<<#var<<":"; \
FOR(i,1,N){ \
std::cerr<<var[i]<<" "; \
} \
std::cerr<<"\n"; \
}while(0)
#define debugMap(variable) \
do { \
std::cerr << #variable << ":\n"; \
for (const auto& pair : variable) { \
std::cerr << " " << pair.first << " => " << pair.second << "\n"; \
} \
std::cerr << std::endl; \
} while (0)
#define lson(var) (var<<1)
#define rson(var) ((var<<1)+1)
#define vecMxx(vec,x) (vec.end()-lower_bound(all(vec),x)) // vec中大于x的数有几个
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef __int128 i128;
typedef pair<ll,ll> pll;
typedef array<ll,3> arr3;
typedef array<ll,2> arr2;
typedef double DB;
typedef long double LD;
typedef pair<DB,DB> pdd;
typedef pair<ll,bool> plb;
constexpr ll MAXN=static_cast<ll>(2e6)+10,INF=static_cast<ll>(1e18)+3;
constexpr ll mod=static_cast<ll>(1e9)+7;
constexpr double eps=1e-8;
ll N,M,Q,X,K,T,A[MAXN];
char S[MAXN];
vector<ll> g[30];
char st[MAXN];
// arr2 lr[MAXN];
/*
*/
inline void solve(){
cin>>N;
ll idx=0;
FOR(i,0,27){g[i].clear();}
FOR(i,0,N-1){
cin>>S[i];
if(g[S[i]-'A'].empty() || i-g[S[i]-'A'].back()>1){
st[++idx]=S[i];
// lr[idx][0]=i;
}else{
// lr[idx][1]=i;
}
g[S[i]-'A'].pb(i);
}
ll ans=1;
if(idx==1){
cout<<1<<"\n";
}else if(idx==2){
cout<<1+min(SZ(g[st[2]-'A']),3)<<"\n";
}else{
cout<<2<<"\n";
}
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
cin>>T;
while(T--){
solve();
}
// solve();
return 0;
}
/*
AC https://ac.nowcoder.com/acm/contest/view-submission?submissionId=77632421
*/
```
:::
## 心路历程（WA，TLE，MLE……）