Submission #761935

#	Time	Username	Problem	Language	Result	Execution time	Memory
761935		GrindMachine	Nafta (COI15_nafta)	C++17	34 / 100	1083 ms	174992 KiB

nafta

// Om Namah Shivaya

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*



*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

struct DSU {
    vector<int> par, rankk, siz;
    vector<int> mnj, mxj, sum;

    DSU() {

    }

    DSU(int n) {
        init(n);
    }

    void init(int n) {
        par = vector<int>(n + 1);
        rankk = vector<int>(n + 1);
        siz = vector<int>(n + 1);
        mnj = vector<int>(n + 1);
        mxj = vector<int>(n + 1);
        sum = vector<int>(n + 1);

        rep(i, n + 1) create(i);
    }

    void create(int u) {
        par[u] = u;
        rankk[u] = 0;
        siz[u] = 1;
        mnj[u] = inf1;
        mxj[u] = -inf1;
        sum[u] = 0;
    }

    int find(int u) {
        if (u == par[u]) return u;
        else return par[u] = find(par[u]);
    }

    bool same(int u, int v) {
        return find(u) == find(v);
    }

    void merge(int u, int v) {
        u = find(u), v = find(v);
        if (u == v) return;

        if (rankk[u] == rankk[v]) rankk[u]++;
        if (rankk[u] < rankk[v]) swap(u, v);

        par[v] = u;
        siz[u] += siz[v];

        amin(mnj[u], mnj[v]);
        amax(mxj[u], mxj[v]);
        sum[u] += sum[v];
    }
};

void solve(int test_case)
{
    ll n,m; cin >> n >> m;
    ll a[n+5][m+5];
    memset(a,-1,sizeof a);

    rep1(i,n){
        rep1(j,m){
            char ch; cin >> ch;
            if(ch != '.'){
                a[i][j] = ch-'0';
            }
            else{
                a[i][j] = -1;
            }
        }
    }

    auto get = [&](ll i, ll j){
        return (i-1)*m + j;
    };

    DSU dsu(n*m+5);
    vector<pll> dir = {{0,1},{1,0}};

    rep1(i,n){
        rep1(j,m){
            ll ind = get(i,j);
            dsu.mnj[ind] = dsu.mxj[ind] = j;
            dsu.sum[ind] = a[i][j];
        }
    }

    rep1(i,n){
        rep1(j,m){
            if(a[i][j] == -1) conts;

            for(auto [dx,dy] : dir){
                ll i2 = i + dx;
                ll j2 = j + dy;

                if(a[i2][j2] != -1){
                    dsu.merge(get(i,j), get(i2,j2));
                }
            }
        }
    }

    vector<pll> here[m+5];

    rep1(i,n){
        rep1(j,m){
            if(a[i][j] == -1) conts;
            ll ind = get(i,j);

            if(dsu.find(ind) == ind){
                ll l = dsu.mnj[ind];
                ll r = dsu.mxj[ind];
                ll w = dsu.sum[ind];
                here[r].pb({l,w});
            }
        }
    }

    ll dp[m+5][m+5];
    memset(dp,-0x3f,sizeof dp);
    dp[0][0] = 0;

    rep1(i,m){        
        vector<ll> pref(i+5);
        for(auto [l,w] : here[i]){
            pref[l] += w;
        }

        rep1(j,i){
            pref[j] += pref[j-1];
        }

        ll tot = pref[i];

        rep1(j,i){
            rep(k,i){
                amax(dp[i][j], dp[k][j-1] + tot);
            }
        }

        rep1(k,i-1){
            rep1(j,k){
                dp[k][j] += pref[k];
            }
        }
    }

    rep1(j,m){
        ll ans = -inf2;

        rep1(i,m){
            amax(ans,dp[i][j]);
        }

        cout << ans << endl;
    }
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

• Subtask 1: 11 / 11
• Subtask 2: 23 / 23
• Subtask 3: 0 / 66