Panasonic Programming Contest 2025（AtCoder Beginner Contest 427）

D - The Simple Game

k<=10, 记忆化搜索,状态数2e5*20

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define pii pair<int,int>
#define ll long long
#define pb push_back
#define ft first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f

#define int long long


const int N = 200010;
vector<int> G[N];
bool a[N][30];
bool vis[N][30];
string s; int mk;

int dfs(int u, int step){
	if(vis[u][step]){
		return !a[u][step];
	}
	vis[u][step] = 1; 

  if(step == 2*mk + 1){
    if(s[u] == 'A') a[u][step] = 1;
	else a[u][step] = 0;
    return !a[u][step];
  }

  a[u][step] = 0;
  for(auto v:G[u]){
	 a[u][step] |= dfs(v, step + 1);
  }
  //cout << (!a[u][step]) << '\n';
  return !a[u][step];
}
void solve(){
    int n, m; cin >> n >> m >> mk;
	for(int i = 1; i <= n; i ++ ) {
		for(int j = 1; j <= 2 * mk + 1; j ++) vis[i][j] = 0;
		G[i].clear();
	}
cin >> s; s = "a" + s;


while(m -- ){
  int u, v; cin >> u >> v; G[u].pb(v);
}

if(dfs(1, 1)){
	cout << "Bob\n"; 
}
else cout << "Alice\n";


}

signed main(){
    std::ios::sync_with_stdio(false);
    int T = 1; cin >> T;
    while(T--){
        solve();
    }
}

E - Wind Cleaning

F - Not Adjacent

折半搜索，双搜
从n/2处断开，前面搜的存在mp里，3* 2^15log(215)

#include<bits/stdc++.h>
using namespace std;
#define endl '\n'
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define pii pair<int,int>
#define ll long long
#define pb push_back
#define ft first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f

//#define int long long

int a[100];
unordered_map<int, ll> mp0;
unordered_map<int, ll> mp1;
int n, m;ll ans;
void dfs1(int step, int cur,  int lst){
    if(step >= n/2) {
        if(lst)
    mp1[cur] ++;
    else mp0[cur] ++;
        return ;
    }

    if(!lst)
    dfs1(step + 1, (1ll*cur + 1ll*a[step]) % m, 1);

    dfs1(step + 1, cur, 0);
}

void dfs2(int step, int cur,  int lst){
    if(step >= n + 1){
        ans += mp0[(m - cur) % m];
        return ;
    }
    if(!lst) dfs2(step + 1, (1ll*cur +  1ll*a[step]) % m, 1);
    dfs2(step + 1, cur, 0);
}
void dfs3(int step, int cur,  int lst){
    if(step >= n + 1){
        ans += mp0[(m - cur) % m] + mp1[(m - cur) % m];
        return ;
    }
    if(!lst) dfs3(step + 1, (1ll*cur +  1ll*a[step]) % m, 1);
    dfs3(step + 1, cur, 0);
}
void solve(){
    cin >> n >> m;
    for(int i = 1; i <= n; i ++) cin >> a[i];
 if(n == 1){
    if(a[1] % m ==0)
        cout << 2 << '\n';
        else cout << 1 << '\n';
        return ;
    }

    dfs1(1, 0, 0);
    dfs2(n/2 + 1, a[n/2], 1);
    dfs3(n/2 + 1, 0, 0);
    cout << ans << '\n';
}

signed main(){
    std::ios::sync_with_stdio(false);
    int T = 1; //cin >> T;
    while(T--){
        solve();
    }
}