Submission #569572

#	Time	Username	Problem	Language	Result	Execution time	Memory
569572		yoavL	Land of the Rainbow Gold (APIO17_rainbow)	C++14	0 / 100	3077 ms	55184 KiB

rainbow

#include "rainbow.h"
#include <iostream>
#include <vector>
#include <algorithm>
#include <string>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <queue>
#include <stack>
#include <bitset>
#include <math.h>
#include <fstream>
#include <iomanip>
#include <functional>
#include <numeric>
#include <chrono>
#include <random>

using namespace std;

using ll = long long;
using ld = long double;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vvvvll = vector<vvvll>;
using vb = vector<bool>;
using vvb = vector<vb>;
using vvvb = vector<vvb>;
using vld = vector<ld>;
using vvld = vector<vld>;
using vstr = vector<string>;
using pll = pair<ll, ll>;
using vpll = vector<pll>;
using vvpll = vector<vpll>;
using pb = pair<bool, bool>;
using vpb = vector<pb>;
using vvpb = vector<vpb>;
using vi = vector<int>;
using vvi = vector<vi>;
using pi = pair<int, int>;
using vpi = vector<pi>;

const ll inf = (ll)2e18;
const ll mod = (ll)(1e9 + 7);

#define FAST        ios_base::sync_with_stdio(0)
#define FASTIN		cin.tie(0)
#define FASTOUT		cout.tie(0)

#define upmin(a, b) (a) = min((a), (b))
#define upmax(a, b) (a) = max((a), (b))

#define pr(x) cout << x << endl
#define prv(v) for(auto it : v) cout << it << " "; cout << endl;
#define prvv(v) for(auto it : v) { for(auto it2 : it) cout << it2 << " "; cout << endl; } cout << endl;
#define spr(x) cout << x << " "

#define DBG_FLAG (1) // Toggle to switch between bebug mode and solution mode
#ifdef DBG_FLAG
#define wpr(x) cout << #x << ": " << (x) << endl;
#define wprv(v) cout << #v << ": " << endl; for(auto it : v) cout << it << " "; cout << endl;
#define wprvv(v) cout << #v << ": " << endl; for(auto it : v) { for(auto it2 : it) cout << it2 << " "; cout << endl; } cout << endl;
#define wspr(x) cout << #x << ": " << x << " "
#endif

#ifndef DBG_FLAG
#define wpr(x)
#define wprv(v)
#define wprvv(v)
#define wspr(x)
#endif

#define rep(i, s, e) for(ll i = s; i < e; i++)
#define all(x) (x).begin(), (x).end()

#define x first
#define y second

/*

Solution:

Walk through all of the nodes adjecent to the river.
That is good enough. Complexity:
O(q * min(M, r*c)) = O(q * M)

Should solve:

Subtask 1, Subtask 3.

Total points: 35

*/

vpll adj;
vpll path;
set<pll> isPath, isAdj;
ll r, c, m, sr, sc;
ll xdir[] = { -1, 1, 0, 0 };
ll ydir[] = { 0, 0, -1, 1 };
ll extxdir[] = { -1, 1, 0, 0, 1, 1, -1, -1 };
ll extydir[] = { 0, 0, -1, 1, 1, -1, 1, -1};


ostream& operator<<(ostream & os, const pll & p)
{
    cout << "{" << p.x << ", " << p.y << "}";
    return os;
}

void init(int R, int C, int Sr, int Sc, int M, char* S)
{
    r = R, c = C, sr = Sr - 1, sc = Sc - 1, m = M + 1;
    
    path.resize(m);
    path[0] = { sr, sc };
    rep(i, 0, m - 1) {
        ll x = path[i].x, y = path[i].y;
        if (S[i] == 'N') x--;
        if (S[i] == 'S') x++;
        if (S[i] == 'W') y--;
        if (S[i] == 'E') y++;
        path[i + 1] = { x, y };
    }
    rep(i, 0, m) {
        ll x = path[i].x, y = path[i].y;
        isPath.insert({ x, y });
    }
    rep(i, 0, m) {
        ll x = path[i].x, y = path[i].y;
        rep(dir, 0, 8) {
            ll nx = x + extxdir[dir];
            ll ny = y + extydir[dir];
            if (nx < 0 || nx >= r || ny < 0 || ny >= c) continue;
            if (isPath.find({ nx, ny }) != isPath.end()) continue;
            adj.push_back({ nx, ny });
            isAdj.insert({ nx, ny });
        }
    }
    sort(adj.begin(), adj.end());
    adj.erase(unique(adj.begin(), adj.end()), adj.end());
}


void dfs_rect(ll x, ll y, ll ar, ll ac, ll br, ll bc, set<pll>& vis)
{
    
    vis.insert({ x, y });
    rep(dir, 0, 4) {
        ll nx = x + xdir[dir];
        ll ny = y + ydir[dir];
        if (nx < ar || nx > br || ny < ac || ny > bc) continue;
        if (isAdj.find({ nx, ny }) == isAdj.end()) continue;
        if (vis.find({ nx, ny }) != vis.end()) continue;
        dfs_rect(nx, ny, ar, ac, br, bc, vis);
    }
}

int colour(int ar, int ac, int br, int bc) {
    ar--, ac--;
    br--, bc--;
    ll components = 0;
    set<pll> vis;
    for (const auto& loc : adj) {
        ll nx = loc.x, ny = loc.y;
        if (nx < ar || nx > br || ny < ac || ny > bc) continue;
        if (vis.find({ nx, ny }) != vis.end()) continue;
        components++;
        dfs_rect(nx, ny, ar, ac, br, bc, vis);
    }
    if (components == 0) {
        ll nx = ar, ny = ac;
        if (isPath.find({ nx, ny }) == isPath.end()) components++;
    }
    return components;
}

/*


6 4 9 4 
3 3 
NWESSWEWS 
2 3 2 3 
3 2 4 4 
5 3 6 4 
1 2 5 3 


*/

• Subtask 1: 0 / 11
• Subtask 2: 0 / 12
• Subtask 3: 0 / 24
• Subtask 4: 0 / 27
• Subtask 5: 0 / 26