제출 #1184982

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1184982		Thunnus	Arcade (NOI20_arcade)	C++20	51 / 100	3 ms	584 KiB

Arcade

#include<bits/stdc++.h>
using namespace std;
using i64 = long long;
#define int i64
#define vi vector<int>
#define vvi vector<vi>
#define vb vector<bool>
#define pii pair<int, int>
#define fi first
#define se second
#define sz(x) (int)(x).size()

inline void solve(){
    int n, m;
    cin >> n >> m;
    vector<pii> a(m); // (x, y), x = a[i], y = t[i], each hand can be descriped by a line heading downwards and connecting points, such that no line exceeds an angle of 45 degrees from the vertical (in both directions).
    for(int i = 0; i < m; i++){
        cin >> a[i].se;
    }
    for(int i = 0; i < m; i++){
        cin >> a[i].fi;
    }
    for(int i = 0; i < m; i++){ // rotate the grid 45 degrees clockwise. Now all lines must connect points where one point is the bottom-left of another. This grid can be viewed as a DAG. 
        a[i] = {a[i].fi - a[i].se, a[i].fi + a[i].se};
    }
    // The question now becomes: How do we decompose this DAG into the minimum number of chains such that each chain follows the edges on that DAG?
    // By Dilworth's theorem the minimum partitioning of a DAG into chains is equal to the size of the maximum antichain.
    // Maximum antichain -> Finding a line connecting the maximum number of points such that the line can only head bottom-right.
    // Finding this line is the equivalent of sorting the points by their y (t) values and finding the longest increasing subsequence by their x (a) values
    sort(a.begin(), a.end(), [&](const pii &lhs, const pii &rhs){return lhs.se < rhs.se;} );

    auto find_lis = [&]() -> int {
        vi lis;
        for(int i = 0; i < m; i++){
            int pos = lower_bound(lis.begin(), lis.end(), a[i].fi) - lis.begin();
            if(pos == sz(lis)){
                lis.emplace_back(a[i].fi);
            }
            else{
                lis[pos] = a[i].fi;
            }
        }
        return sz(lis);
    };

    cout << find_lis() << "\n";
    return;
}

signed main(){
    ios_base::sync_with_stdio(false); cin.tie(0);
    int t = 1;
    //cin >> t;
    while(t--){
        solve();
    }
    return 0;
}

• Subtask 1: 7 / 7
• Subtask 2: 11 / 11
• Subtask 3: 12 / 12
• Subtask 4: 21 / 21
• Subtask 5: 0 / 14
• Subtask 6: 0 / 20
• Subtask 7: 0 / 15