Submission #832506

#	Time	Username	Problem	Language	Result	Execution time	Memory
832506		maomao90	Sky Walking (IOI19_walk)	C++17	0 / 100	49 ms	43940 KiB

walk

// I can do all things through Christ who strenghthens me
// Philippians 4:13

#include "bits/stdc++.h"
#include "walk.h"
using namespace std;

#define REP(i, j, k) for (int i = j; i < (k); i++)
#define RREP(i, j, k) for (int i = j; i >= (k); i--)

template <class T>
inline bool mnto(T &a, const T b) {return a > b ? a = b, 1 : 0;}
template <class T>
inline bool mxto(T &a, const T b) {return a < b ? a = b, 1 : 0;}

typedef long long ll;
typedef long double ld;
#define FI first
#define SE second
typedef pair<int, int> ii;
typedef pair<ll, ll> pll;
#define ALL(x) x.begin(), x.end()
#define SZ(x) (int) x.size()
#define pb push_back
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<ll> vll;
typedef tuple<int, int, int> iii;
typedef tuple<ll, int, int> lll;
typedef vector<iii> viii;

#ifndef DEBUG
#define cerr if (0) cerr
#endif

const int INF = 1000000005;
const ll LINF = 1000000000000000005;
const int MAXN = 100005;

int n, m;
vi vy[MAXN];
map<int, int> mp[MAXN];
vll d[MAXN];
vi prv[MAXN], nxt[MAXN];
priority_queue<lll, vector<lll>, greater<lll>> pq;

ll min_distance(vi x, vi h, vi l, vi r, vi y, int s, int g) {
    n = SZ(x), m = SZ(l);
    REP (i, 0, m) {
        REP (p, l[i], r[i] + 1) {
            if (h[p] < y[i]) {
                continue;
            }
            vy[p].pb(i);
        }
    }
    REP (i, 0, n) {
        d[i] = vll(SZ(vy[i]), LINF);
        prv[i] = vi(SZ(vy[i]), -1);
        nxt[i] = vi(SZ(vy[i]), INF);
        sort(ALL(vy[i]), [&] (int l, int r) {
                return y[l] < y[r];
                });
        REP (j, 0, SZ(vy[i])) {
            mp[i][y[vy[i][j]]] = j;
        }
    }
    REP (i, 0, m) {
        int tprv = -1;
        REP (p, l[i], r[i] + 1) {
            if (h[p] < y[i]) {
                continue;
            }
            prv[p][mp[p][y[i]]] = tprv;
            if (tprv != -1) {
                nxt[tprv][mp[tprv][y[i]]] = p;
            }
            tprv = p;
        }
    }
    REP (j, 0, SZ(vy[s])) {
        d[s][j] = y[vy[s][j]];
        pq.push({d[s][j], s, j});
    }
    while (!pq.empty()) {
        auto [ud, up, ui] = pq.top(); pq.pop();
        if (ud != d[up][ui]) {
            continue;
        }
        int id = vy[up][ui];
        for (int j : {max(0, (int) ui - 1), min(SZ(vy[up]), (int) ui + 2)}) {
            if (mnto(d[up][j], ud + abs(y[id] - y[vy[up][j]]))) {
                pq.push({d[up][j], up, j});
            }
        }
        for (int p : {max(l[id], prv[up][ui]), min(r[id], nxt[up][ui])}) {
            int j = mp[p][y[id]];
            if (mnto(d[p][j], ud + abs(x[p] - x[up]))) {
                pq.push({d[p][j], p, j});
            }
        }
    }
    ll ans = LINF;
    REP (j, 0, SZ(vy[g])) {
        mnto(ans, d[g][j] + y[vy[g][j]]);
    }
    if (ans == LINF) {
        return -1;
    }
	return ans;
}

• Subtask 1: 0 / 10
• Subtask 2: 0 / 14
• Subtask 3: 0 / 15
• Subtask 4: 0 / 18
• Subtask 5: 0 / 43