답안 #814444

# 제출 시각 아이디 문제 언어 결과 실행 시간 메모리
814444 2023-08-08T07:30:24 Z becaido 분수 공원 (IOI21_parks) C++17
0 / 100
3 ms 4948 KB
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,popcnt,sse4,abm")
#include <bits/stdc++.h>
using namespace std;

#ifndef WAIMAI
#include "parks.h"
#endif

#ifdef WAIMAI
#define debug(HEHE...) cout << "[" << #HEHE << "] : ", dout(HEHE)
void dout() {cout << '\n';}
template<typename T, typename...U>
void dout(T t, U...u) {cout << t << (sizeof...(u) ? ", " : ""), dout(u...);}
#else
#define debug(...) 7122
#endif

#define ll long long
#define Waimai ios::sync_with_stdio(false), cin.tie(0)
#define FOR(x,a,b) for (int x = a, I = b; x <= I; x++)
#define pb emplace_back
#define F first
#define S second

#ifdef WAIMAI
void build(vector<int> u, vector<int> v, vector<int> a, vector<int> b);
#endif

const int dx[4] = {-1, 0, 0, 1};
const int dy[4] = {0, -1, 1, 0};
const int SIZE = 2e5 + 5;

vector<int> adj[SIZE];
map<pair<int, int>, int> mp;
int deg[SIZE], pa[SIZE];
queue<int> q;

int construct_roads(vector<int> x, vector<int> y) {
    int n = x.size();
    FOR (i, 0, n - 1) mp[{x[i], y[i]}] = i;
    FOR (i, 0, n - 1) FOR (j, 0, 3) {
        int tx = x[i] + 2 * dx[j];
        int ty = y[i] + 2 * dy[j];
        if (mp.count({tx, ty})) adj[i].pb(mp[{tx, ty}]);
    }
    q.push(0);
    fill(pa, pa + n, -1);
    while (q.size()) {
        int pos = q.front();
        q.pop();
        for (int np : adj[pos]) if (np != 0 && pa[np] == -1) {
            pa[np] = pos;
            deg[pos]++;
            q.push(np);
        }
    }
    FOR (i, 1, n - 1) if (pa[i] == -1) return 0;
    FOR (i, 0, n - 1) if (!deg[i]) q.push(i);
    vector<int> u, v, a, b;
    while (q.size()) {
        int pos = q.front();
        q.pop();
        if (pos == 0) break;
        u.pb(pos), v.pb(pa[pos]);
        if (--deg[pa[pos]] == 0) q.push(pa[pos]);
    }
    a.resize(n - 1), b.resize(n - 1);
    FOR (i, 0, n - 2) if (x[u[i]] == 2 && x[v[i]] == 2) {
        a[i] = x[u[i]] - 1;
        b[i] = y[u[i]];
    }
    FOR (i, 0, n - 2) if (x[u[i]] == 6 && x[v[i]] == 6) {
        a[i] = x[u[i]] + 1;
        b[i] = y[u[i]];
    }
    set<pair<int, int>> s;
    FOR (i, 0, n - 2) {
        int lx = min(x[u[i]], x[v[i]]), rx = max(x[u[i]], x[v[i]]);
        if (lx == 2 && rx == 4) {
            a[i] = 3, b[i] = y[u[i]] + 1;
            s.emplace(a[i], b[i]);
        }
    }
    FOR (i, 0, n - 2) if (x[u[i]] == 4 && x[v[i]] == 4) {
        a[i] = x[u[i]] - 1, b[i] = y[u[i]];
        if (s.count({a[i], b[i]})) a[i] += 2;
        s.emplace(a[i], b[i]);
    }
    FOR (i, 0, n - 2) {
        int lx = min(x[u[i]], x[v[i]]), rx = max(x[u[i]], x[v[i]]);
        if (lx == 4 && rx == 6) {
            a[i] = 5, b[i] = y[u[i]] + 1;
            if (s.count({a[i], b[i]})) b[i] -= 2;
            s.emplace(a[i], b[i]);
        }
    }
    build(u, v, a, b);
    return 1;
}

/*
in1
5
4 4
4 6
6 4
4 2
2 4
out1
1
4
0 2 5 5
0 1 3 5
3 0 5 3
4 0 3 3

in2
2
2 2
4 6
out2
0
*/

#ifdef WAIMAI
static void check(bool cond, string message) {
	if (!cond) {
		printf("%s\n", message.c_str());
		fclose(stdout);
		exit(0);
	}
}

static int n;
static bool build_called;
static int m;
static vector<int> _u, _v, _a, _b;

void build(vector<int> u, vector<int> v, vector<int> a, vector<int> b) {
	check(!build_called, "build is called more than once");
	build_called = true;
	m = u.size();
	check(int(v.size()) == m, "u.size() != v.size()");
	check(int(a.size()) == m, "u.size() != a.size()");
	check(int(b.size()) == m, "u.size() != b.size()");
	_u = u;
	_v = v;
	_a = a;
	_b = b;
}

int main() {
	assert(scanf("%d", &n) == 1);
	vector<int> x(n), y(n);
	for (int i = 0; i < n; i++) {
		assert(scanf("%d%d", &x[i], &y[i]) == 2);
	}
	fclose(stdin);

	build_called = false;
	const int possible = construct_roads(x, y);

	check(possible == 0 || possible == 1, "Invalid return value of construct_roads()");
	if (possible == 1) {
		check(build_called, "construct_roads() returned 1 without calling build()");
	} else {
		check(!build_called, "construct_roads() called build() but returned 0");
	}

	printf("%d\n", possible);
	if (possible == 1) {
		printf("%d\n", m);
		for (int j = 0; j < m; j++) {
			printf("%d %d %d %d\n", _u[j], _v[j], _a[j], _b[j]);
		}
	}
	fclose(stdout);
	return 0;
}
#endif
# 결과 실행 시간 메모리 Grader output
1 Correct 3 ms 4948 KB Output is correct
2 Incorrect 3 ms 4948 KB b[0] = 4 is not an odd integer
3 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Correct 3 ms 4948 KB Output is correct
2 Incorrect 3 ms 4948 KB b[0] = 4 is not an odd integer
3 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Correct 3 ms 4948 KB Output is correct
2 Incorrect 3 ms 4948 KB b[0] = 4 is not an odd integer
3 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Correct 3 ms 4948 KB Output is correct
2 Incorrect 3 ms 4948 KB b[0] = 4 is not an odd integer
3 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Correct 3 ms 4948 KB Output is correct
2 Incorrect 3 ms 4948 KB b[0] = 4 is not an odd integer
3 Halted 0 ms 0 KB -
# 결과 실행 시간 메모리 Grader output
1 Correct 3 ms 4948 KB Output is correct
2 Incorrect 3 ms 4948 KB b[0] = 4 is not an odd integer
3 Halted 0 ms 0 KB -