Submission #1191404

#	Time	Username	Problem	Language	Result	Execution time	Memory
1191404		Ardelion	Jakarta Skyscrapers (APIO15_skyscraper)	C++20	36 / 100	74 ms	24136 KiB

skyscraper

#include <bits/stdc++.h>
//#pragma GCC optimize("Ofast")
//#pragma GCC optimize("O3")
//#pragma GCC target("avx2")

using namespace std;

//#define int          long long
#define ll           long long
#define X            first
#define Y            second
#define lc           (id<<1)
#define rc           (lc|1)
#define mid          ((l+r+1)>>1)
#define pb           push_back
#define lb           lower_bound
#define ub           upper_bound
#define mp           make_pair
#define sep          ' '
#define endl         "\n"
#define migmig       ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define FileIO       freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
#define all(x)       x.begin(), x.end()
#define ret(x)       {cout << x << endl; return;}

typedef pair<int, int> pii;
typedef pair<ll , ll > pll;

const int N     = 3e3+13;
const int LOG   = 23;
const int MOD   = 1e9 + 7; //998244353; //1e9+9;
const int INF   = 1e9; //1e18;
const int dx[4] = {1, -1, 0,  0};
const int dy[4] = {0,  0, 1, -1};

ll md(ll x) {x%=MOD; return (x < 0 ? x+MOD : x);}
ll GCD(ll _a, ll _b) { return (!_b ? _a : GCD(_b, _a % _b)); }
ll POW(ll _a, ll _b) { return !_b ? 1 : ((_b & 1 ? _a : 1) * POW(_a * _a % MOD, _b / 2)) % MOD;}

int n, m, s, f;
int dist[N][N], P[N];
vector<int> b[N];

void BFS() {
    deque<pii> q;
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            dist[i][j] = INF;
        }
    }
    dist[s][0] = 0;
    q.push_back(mp(s, 0));
    pii p;
    while(!q.empty()) {
        p = q.front();
        q.pop_front();
        for(int d : b[p.X]) {
            if(dist[p.X][d] > dist[p.X][p.Y]) {
                dist[p.X][d] = dist[p.X][p.Y];
                q.push_front(mp(p.X, d));
            }
        }
        if(p.X - P[p.Y] >= 0 && dist[p.X-P[p.Y]][p.Y] > dist[p.X][p.Y] + 1) {
            dist[p.X-P[p.Y]][p.Y] = dist[p.X][p.Y] + 1;
            q.push_back(mp(p.X-P[p.Y], p.Y));
        }
        if(p.X + P[p.Y] < n && dist[p.X+P[p.Y]][p.Y] > dist[p.X][p.Y] + 1) {
            dist[p.X+P[p.Y]][p.Y] = dist[p.X][p.Y] + 1;
            q.push_back(mp(p.X+P[p.Y], p.Y));
        }
    }
}

void F() {
    cin >> n >> m;
    int x;
    for(int i = 0; i < m; i++) {
        cin >> x >> P[i];
        if(i == 0) s = x;
        else if(i == 1) f = x;
        b[x].pb(i);
    }
    BFS();
    cout << (dist[f][1] == INF ? -1 : dist[f][1]);
}

int32_t main() {
    migmig;
    //PREP();
    int t=1;
    //cin >> t;
    while(t--)
        F();
    return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 12 / 12
• Subtask 3: 14 / 14
• Subtask 4: 0 / 21
• Subtask 5: 0 / 43