Submission #336790

#TimeUsernameProblemLanguageResultExecution timeMemory
336790super_j6Abduction 2 (JOI17_abduction2)C++14
27 / 100
5092 ms14264 KiB
#include <iostream> #include <cstdio> #include <algorithm> #include <array> #include <unordered_map> using namespace std; #define endl '\n' #define ll long long #define pi pair<int, int> #define f first #define s second typedef array<int, 2> T; T operator+(T x, T y){ return {x[0] + y[0], x[1] + y[1]}; } const int mxn = 50000, k = 4; const T d[k] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}}; int n[2], q; int a[2][mxn]; unordered_map<int, int> dp[k][mxn]; bool f(T x, int y){ if(x[y & 1] < 0 || x[y & 1] >= n[y & 1]) return 0; return 1; } int sol(T x, int y){ int ret = 0, s = 0; for(; f(x, y) && a[!(y & 1)][x[!(y & 1)]] > a[y & 1][x[y & 1]]; x = x + d[y]) s++; if(!f(x, y)) return s; if(dp[y][x[0]].count(x[1])) return dp[y][x[0]][x[1]] + s; for(int i = 0; i < 2; i++){ int z = (y + 2 * i + 1) % k; ret = max(ret, sol(x + d[z], z)); } dp[y][x[0]][x[1]] = ++ret; return ret + s; } int main(){ ios::sync_with_stdio(0); cin.tie(0); cin >> n[0] >> n[1] >> q; for(int i = 0; i < 2; i++) for(int j = 0; j < n[i]; j++){ cin >> a[i][j]; } while(q--){ T x; for(int i = 0; i < 2; i++) cin >> x[i], x[i]--; int ret = 0; for(int i = 0; i < k; i++) ret = max(ret, sol(x + d[i], i)); cout << ret << endl; } return 0; }
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