提出 #60793631

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ソースコード 拡げる

// 二次元グリッドの問題 priority_queueの利用
#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#include <atcoder/all>
using namespace atcoder;
using ll=long long;
using P=pair<int,int>;
using IP=pair<ll,P>;
using mint=modint998244353;

const int di[4] = {-1, 0, 1, 0};
const int dj[4] = { 0, 1, 0,-1};

int main(){
    int h,w,x;cin >> h >> w >> x;
    int si,sj; cin >> si >> sj;
    --si, --sj;
    vector<vector<ll>> s(h,vector<ll>(w));
    rep(i,h) rep(j,w) cin >> s[i][j];

    ll now = s[si][sj];
    vector visited(h,vector<bool>(w));
    priority_queue<IP,vector<IP>,greater<IP>> p;
    auto nighbour=[&](int i, int j){
        rep(k,4) {
            int ni = i + di[k];
            int nj = j + dj[k];
            if (ni < 0 || ni >= h || nj < 0 || nj >= w) continue;
            if (visited[ni][nj]) continue;
            visited[ni][nj] = true;
            p.push(IP(s[ni][nj], P(ni,nj)));
        }
    };
    visited[si][sj] = true;
    nighbour(si,sj);
    while (!p.empty()) {
        auto [power, pos] = p.top(); p.pop();
        if (power >= (now + x - 1) / x) break;
        int i = pos.first;
        int j = pos.second;
        now += power;
        nighbour(i,j);
    }
    cout << now << endl;

    return 0;
}

提出情報

ジャッジ結果