제출 #637756

#제출 시각아이디문제언어결과실행 시간메모리
637756ksu2009enT-Covering (eJOI19_covering)C++14
35 / 100
215 ms16576 KiB
#include <iostream> #include <vector> #include <string> #include <math.h> #include <cmath> #include <iomanip> #include <cstdio> #include <algorithm> #include <map> #include <set> #include <queue> #include <stack> #include <deque> #include <bitset> #include <cstring> using namespace std; typedef long long ll; vector<pair<ll, ll>> state(ll x, ll y, ll ok){ if(ok == 1) return {{x, y}, {x-1, y}, {x, y - 1}, {x, y + 1}}; if(ok == 2) return {{x, y}, {x-1, y}, {x + 1, y}, {x, y + 1}}; if(ok == 3) return {{x, y}, {x+1, y}, {x, y - 1}, {x, y + 1}}; return {{x, y}, {x-1, y}, {x + 1, y}, {x, y - 1}}; } ll n, m; vector<vector<ll>>used; ll check(ll x, ll y, vector<vector<ll>>&a, ll ok){ auto d = state(x, y, ok); ll ans = 0; for(auto i: d){ if(i.first <= 0 || i.second <= 0 || i.first > n || i.second > m) return -1e9; if(i.first != x || i.second != y){ if(used[i.first][i.second] == 1) return -1e9; } ans += a[i.first][i.second]; } return ans; } int main(){ cin >> n >> m; used.resize(n + 1, vector<ll>(m + 1)); vector<vector<ll>>a(n + 1, vector<ll>(m + 1)); for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) cin >> a[i][j]; ll k; cin >> k; vector<pair<ll, ll>> d(k); for(int i = 0; i < k; i++){ cin >> d[i].first >> d[i].second; d[i].first++, d[i].second++; used[d[i].first][d[i].second] = 1; } sort(d.begin(), d.end()); vector<vector<ll>>dp(k + 1, vector<ll>(5)); for(int i =0 ; i < k; i++) for(int j = 1; j <= 4; j++) dp[i][j] = -1e9; for(int i = 1; i <= 4; i++){ dp[0][i] = check(d[0].first, d[0].second, a, i); } for(int i = 1; i< k; i++){ for(int j = 1; j <= 4; j++){ for(int last = 1; last <= 4; last++){ auto v1 = state(d[i].first, d[i].second, j); auto v2 = state(d[i - 1].first, d[i - 1].second, last); set<pair<ll, ll>> st; bool ok = false; for(auto l: v1){ if(st.count(l)) ok = true; st.insert(l); } for(auto l: v2){ if(st.count(l)) ok = true; st.insert(l); } if(ok) // overlaping continue; dp[i][j] = max(dp[i][j], dp[i - 1][last] + check(d[i].first, d[i].second, a, j)); } } } ll mx = -1; for(int j = 1; j <= 4; j++) mx = max(mx, dp[k - 1][j]); if(mx < 0) cout << "No" << endl; else cout << mx << endl; return 0; }
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