Submission #637757

#	Time	Username	Problem	Language	Result	Execution time	Memory
637757		ksu2009en	T-Covering (eJOI19_covering)	C++14	0 / 100	2 ms	340 KiB

covering

#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());
    
    set<ll>st;
    for(auto i: d)
        st.insert(i.second);
    
    if(st.size() == 1){
        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;
    }
    return 0;
}

• Subtask 1: 0 / 5
• Subtask 2: 0 / 10
• Subtask 3: 0 / 10
• Subtask 4: 0 / 10
• Subtask 5: 0 / 15
• Subtask 6: 0 / 20
• Subtask 7: 0 / 30