제출 #637759

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
637759		ksu2009en	T-Covering (eJOI19_covering)	C++14	15 / 100	1086 ms	468 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;
}
ll t[10007];
vector<pair<ll, ll>> d;
vector<vector<ll>>a;
 
ll res = -1e9;
 
void put(ll pos, ll val, ll& mx){
    t[pos] = val;
    
    if(pos == mx){
        set<pair<int, int>> st;
        
        bool ok = false;
        
        ll sum = 0;
        
        for(int i = 1; i <= pos; i++){
            if(st.count({d[i - 1].first, d[i - 1].second})){
                ok = true;
                break;
            }
            st.insert({d[i - 1].first, d[i - 1].second});
            if(t[i] == 1){
                if(st.count({d[i - 1].first - 1, d[i - 1].second}) || st.count({d[i - 1].first, d[i - 1].second - 1}) || st.count({d[i - 1].first, d[i - 1].second + 1})){
                    ok = true;
                    break;
                }
                st.insert({d[i - 1].first - 1, d[i - 1].second});
                st.insert({d[i - 1].first, d[i - 1].second - 1});
                st.insert({d[i - 1].first, d[i - 1].second + 1});
                
            }
            else if(t[i] == 2){
                if(st.count({d[i - 1].first - 1, d[i - 1].second}) || st.count({d[i - 1].first + 1, d[i - 1].second}) || st.count({d[i - 1].first, d[i - 1].second + 1})){
                    ok = true;
                    break;
                }
                st.insert({d[i - 1].first - 1, d[i - 1].second});
                st.insert({d[i - 1].first + 1, d[i - 1].second});
                st.insert({d[i - 1].first, d[i - 1].second + 1});
            }
            else if(t[i] == 3){
                if(st.count({d[i - 1].first, d[i - 1].second - 1}) || st.count({d[i - 1].first, d[i - 1].second + 1}) || st.count({d[i - 1].first + 1, d[i - 1].second})){
                    ok = true;
                    break;
                }
                st.insert({d[i - 1].first, d[i - 1].second - 1});
                st.insert({d[i - 1].first, d[i - 1].second + 1});
                st.insert({d[i - 1].first + 1, d[i - 1].second});
            }
            else{
                if(st.count({d[i - 1].first - 1, d[i - 1].second}) || st.count({d[i - 1].first + 1, d[i - 1].second}) || st.count({d[i - 1].first, d[i - 1].second - 1})){
                    ok = true;
                    break;
                }
                st.insert({d[i - 1].first - 1, d[i - 1].second});
                st.insert({d[i - 1].first + 1, d[i - 1].second});
                st.insert({d[i - 1].first, d[i - 1].second - 1});
            }
            
            if(check(d[i - 1].first, d[i - 1].second, a, t[i]) < 0){
                ok = true;
                break;
            }
            
            sum += check(d[i - 1].first, d[i - 1].second, a, t[i]);
        }
        
        if(!ok){
            res = max(res, sum);
        }
    }
    else
        for(int i = 1; i <= 4; i++)
            put(pos + 1, i, mx);
}

int main(){
    cin >> n >> m;
    
    used.resize(n + 1, vector<ll>(m + 1));
    
    a.resize(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;
    
    d.resize(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());
    
    if(k <= 10){
        for(int i = 1; i <= 4; i++)
            put(1, i, k);
                
        if(res == -1e9)
            cout << "No" << endl;
        else
            cout << res << endl;
        return 0;
    }
    
    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;
}

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