Submission #223874

#TimeUsernameProblemLanguageResultExecution timeMemory
223874super_j6Stray Cat (JOI20_stray)C++14
100 / 100
106 ms16796 KiB
#include "Anthony.h" #include <iostream> #include <cstdio> #include <algorithm> #include <functional> #include <vector> #include <queue> using namespace std; #define endl '\n' #define ll long long #define pi pair<int, int> #define f first #define s second #define vi vector<int> const int maxn = 20000, k = 6; int d[maxn]; vi graph[maxn]; queue<int> q; int s[k] = {0, 0, 1, 0, 1, 1}; vi ans; vi Mark(int n, int m, int a, int b, vi u, vi v){ ans.resize(m); for(int i = 0; i < m; i++){ graph[u[i]].push_back(i); graph[v[i]].push_back(i); } d[0] = 1; if(a > 2){ q.push(0); while(!q.empty()){ int c = q.front(); q.pop(); for(int i : graph[c]){ int j = u[i] ^ v[i] ^ c; if(!d[j]){ d[j] = d[c] + 1; q.push(j); } if(d[j] == d[c] || d[j] == d[c] + 1){ ans[i] = d[c] % 3; } } } }else{ function<void(int, int)> dfs = [&](int c, int p){ for(int i : graph[c]){ int j = u[i] ^ v[i] ^ c; if(j != p){ ans[i] = s[d[c]]; d[j] = graph[j].size() > 2 ? (ans[i] ^ 1) << 1 : (d[c] + 1) % k; dfs(j, c); } } }; dfs(0, -1); } return ans; }
#include "Catherine.h" #include <iostream> #include <cstdio> #include <algorithm> #include <vector> using namespace std; #define endl '\n' #define pi pair<int, int> #define f first #define s second #define vi vector<int> const int k = 6; int a, b; int prv = -1; bool f = 0; int s[k] = {0, 0, 1, 0, 1, 1}; vi t; void Init(int A, int B){ a = A, b = B; } int Move(vi v){ if(a > 2){ if(~prv) v[prv]++; for(int i = 0; i < 3; i++){ if(v[i] && v[(i + 1) % 3]) return prv = i; } for(int i = 0; i < 3; i++){ if(v[i]) return prv = i; } }else{ if(f){ if(v[0] + v[1] > 1) v[prv]++; return prv = (v[0] == 1) ^ 1; } if(!~prv){ if(v[0] + v[1] != 2){ f = 1; return prv = (v[0] == 1) ^ 1; } prv = (v[0] > 0) ^ 1; t.push_back((v[0] > 1) ^ 1); t.push_back(prv); return prv; } if(v[0] + v[1] == 0){ f = 1; return -1; } if(v[0] + v[1] > 1){ f = 1; return !v[prv] ? -1 : prv ^= 1; } t.push_back((v[0] == 1) ^ 1); if(t.size() == k - 1){ for(int i = 0; i < k; i++){ f = 1; for(int j = 0; j < k - 1; j++) f &= s[(i + j) % k] == t[j]; if(f) return -1; } f = 1; } return prv = (v[0] == 1) ^ 1; } return -2; }
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