Submission #906699

#TimeUsernameProblemLanguageResultExecution timeMemory
906699vjudge1메기 농장 (IOI22_fish)C++17
9 / 100
131 ms17636 KiB
#include "fish.h"

#pragma GCC target ("avx2")
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")

#include<bits/stdc++.h>
#include<math.h>
using namespace std;

typedef long long int ll;
typedef long double ld;
typedef pair<ll, ll> pl;
typedef vector<ll> vl;
#define FD(i, r, l) for(ll i = r; i > (l); --i)

#define K first
#define V second
#define G(x) ll x; cin >> x;
#define GD(x) ld x; cin >> x;
#define GS(s) string s; cin >> s;
#define EX(x) { cout << x << '\n'; exit(0); }
#define A(a) (a).begin(), (a).end()
#define F(i, l, r) for (ll i = l; i < (r); ++i)

#define NN 310

ll grid[NN][NN];
ll n;

ll dp1[NN][NN];
ll noreq(ll, ll); // no request fishies, so len == prev wall. 

ll dp2[NN][NN];
ll req(ll, ll); 

// previous column did NOT request any right fishies here
// so we simply have a free wall to attach to 
ll noreq(ll i, ll len) {
    // assert(len == 0 or len == n);
    if (i == n) return 0;
    auto &DP = dp1[i][len];
    if (!~DP) {
        DP = noreq(i+1, n); // no restrictions so build big wall.

        F(nlen, 0, n+1) DP = max(DP, req(i, nlen)); // basically can set up any wall here before taking right fishes

        ll lsum = 0;
        F(j, 0, len) {
            lsum += grid[i][j];
        }
        DP = max(DP, lsum + noreq(i+1, 0));
        {
            ll tsum = lsum;
            F(j, 0, len) {
                tsum -= grid[i][j];
                DP = max(DP, tsum + noreq(i+1, j+1));
            }
        }


        F(covering, len, n) {
            // not lsum naymore jsut cum sum
            lsum += grid[i][covering];
            DP = max(DP, lsum + req(i+1, covering + 1));
        }

    }   
    return DP;   
}


// previous column requested right fishes here;
// so len == min bound on wall (we cannot take anything below len)

ll req(ll i, ll len) {
    if (i == n) return len == 0 ? 0 : -9e18; // i should need 0 fishes here
    auto &DP = dp2[i][len];
    if (!~DP) {
        // note that we cannot request any left fishes here.
        DP = noreq(i+1, n);  // if we don't request ANY fish here, just go up to max 

        ll sm = 0;
        F(covering, len, n) {
            sm += grid[i][covering];
            DP = max(DP, sm + req(i+1, covering + 1));
        }
    }   
    return DP;
}

long long max_weights(int N, int M, std::vector<int> X, std::vector<int> Y,
                      std::vector<int> W) {
    ll rans;

    if (N <= 300) {
        memset(dp1, -1, sizeof dp1);
        memset(dp2, -1, sizeof dp2);
        
        n = N;
        memset(grid, 0, sizeof grid);
        F(i, 0, M) grid[X[i]][Y[i]] = W[i];
        
        rans = noreq(0, 0);          
    }
    constexpr ll DEBUG = 01;
    if (N <= 300 and !DEBUG) return rans;

    bool case1 = 1;
    bool case2 = 1;
    bool case3 = 1;
    F(i, 0, M) case1 &= X[i]%2 == 0;
    F(i, 0, M) case2 &= X[i] <= 1;
    F(i, 0, M) case3 &= Y[i] == 0;
    
    if (case1) {
        return accumulate(A(W), 0ll);
    } else if (case2) {
        ll c[2] = {};
        map<pl, ll> points;
        F(i, 0, M) {
            c[X[i]] += W[i];
            points[{X[i], Y[i]}] = W[i];
        }
        
        if (N == 2) {
            return max(c[0], c[1]);
        }
        ll tans = max(c[0], c[1]);
        ll tsum = c[1];
        F(i, 0, N) {
            tsum -= points[{1, i}];
            tsum += points[{0, i}];
            tans = max(tans, tsum);
        }

        return tans;
    } else if (case3) {
        vl grid(n);
        F(i, 0, M) grid[X[i]] = W[i];
        vector<vl> dp(n+10, vl(3, -1));
        auto rec = [&](auto &&self, ll i, ll f) -> ll {
            if (i > n) return -1e18;
            if (i >= n) return 0;
            auto &DP = dp[i][f];
            if (!~DP) {
                DP = self(self, i+1, 1);
                if (f) DP = max(DP, grid[i] + self(self, i+1, 0));
                DP = max(DP, grid[i] + self(self, i+2, 1));
            }
            return DP;
        };
        // cout << rec(rec, 0, 0) << ' ' << rans << endl;
        // assert(rec(rec, 0, 0) == rans);
        return rec(rec, 0, 0);
    }

    return -1;
}
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