2018 Benelux Algorithm Programming Contest (BAPC 18)

目录

• Contest Info
• Solutions
  • A A Prize No One Can Win
  • B Birthday Boy
  • C Cardboard Container
  • D Driver Disagreement
  • E Entirely Unsorted Sequences
  • F Financial Planning
  • G Game Night
  • H Harry the Hamster
  • I In Case of an Invasion, Please. . .
  • J Janitor Troubles
  • K Kingpin Escape

Contest Info

---

[Practice Link](https://codeforc.es/gym/102007)

Solved	A	B	C	D	E	F	G	H	I	J	K
8/11	O	O	O	-	-	O	O	-	O	O	O

• O 在比赛中通过
• Ø 赛后通过
• ! 尝试了但是失败了
• - 没有尝试

Solutions

---

A A Prize No One Can Win

签到。

view code

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 10;
int n, a[N], x;

int main() {
	while (scanf("%d%d", &n, &x) != EOF) {
		for (int i = 1; i <= n; ++i) scanf("%d", a + i);
		sort(a + 1, a + 1 + n);
		int res = 1;
		for (int i = 2; i <= n; ++i) {
			if (a[i] + a[i - 1] <= x) {
				++res;
			} else break;
		}
		printf("%d\n", res);
	}
	return 0;
}

B Birthday Boy

题意：

给出\(n\)个生日，现在要找个日期，使得在它前面离它最近的生日和它相距的天数最大，如果有多个，输出和\(10-27\)相距天数最大的。

思路：

题意读清楚就可以了，枚举每一天。

view code

#include <bits/stdc++.h>
using namespace std;

using pII = pair <int, int>;
#define fi first
#define se second
int n; pII a[110]; char s[110];
int mon[] = {
	0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31
};
bool operator < (pII a, pII b) {
	if (a.fi < b.fi || (a.fi == b.fi && a.se <= b.se)) {
		return true;
	}
	return false;
}
int dis(pII a, pII b) {
	if (a < b) {
		if (a.fi == b.fi) return b.se - a.se;
		int res = mon[a.fi] - a.se;
		for (int i = a.fi + 1; i < b.fi; ++i) res += mon[i];
		res += b.se;
		return res;
	} else {
		int res = mon[a.fi] - a.se;
		for (int i = a.fi + 1; i <= 12; ++i) res += mon[i];
		for (int i = 1; i < b.fi; ++i) res += mon[i];
		res += b.se;
		return res;
	}
}

int main() {
	while (scanf("%d", &n) != EOF) {
		for (int i = 1; i <= n; ++i) scanf("%s %02d-%02d", s, &a[i].fi, &a[i].se);
		sort(a + 1, a + 1 + n, [&](pII x, pII y) {
			if (x.fi != y.fi) return x.fi < y.fi;
			return x.se < y.se;
		});
		pII res = pII(-1, -1); int Max = -1;
		a[0] = a[n];
		int pos = 0;
		for (int i = 1; i <= 12; ++i) {
			for (int j = 1; j <= mon[i]; ++j) {
				pII t = pII(i, j);
				while (pos < n && a[pos + 1] < t) ++pos;
				if (dis(a[pos], t) > Max) {
					Max = dis(a[pos], t);
					res = t;
				} else if (dis(a[pos], t) == Max) {
					if (dis(pII(10, 28), t) < dis(pII(10, 28), res)) {
						res = t;
					}
				}
			}
		}
		printf("%02d-%02d\n", res.fi, res.se);
	}
	return 0;
}

C Cardboard Container

题意：

给出一个立方体的体积\(V\)，算表面积

思路：

枚举长宽高，长宽高是\(V\)的因子，枚举量不大。

view code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
void getfac(vector <int> &vec, int n) {
	vec.clear();
	for (int i = 1; 1ll * i * i <= n; ++i) {
		if (n % i == 0) {
			vec.push_back(i);
			if (i * i != n)
			vec.push_back(n / i);
		}
	}
}

int main() {
	int n;
	while (scanf("%d", &n) != EOF) {
		vector <int> vec_a;
		getfac(vec_a, n);
		ll res = 1e18;
		for (auto a : vec_a) {
			vector <int> vec_b;
			getfac(vec_b, n / a);
			for (auto &b : vec_b) {
				int c = n / a / b;
				res = min(res, 2ll * (a * b + a * c + b * c));
			}
		}
		printf("%lld\n", res);
	}
	return 0;
}

D Driver Disagreement

题意：

给出\(n\)个点的图，每个点都有两条边，并且每个点有一个标记\(1\)或者\(0\)

现在两个人在同一点，但是他们不知道自己在哪一点，一个人觉得他们在\(A\)，另一个人觉得他们在\(B\)。

然后他们所在的位置是\(A\)或者\(B\)

现在需要设计一种路线，使得沿着路线走，如果从\(A\)出发走到一个点和从\(B\)出发走到一个点那个点的标记不一样，那么它们就知道谁对谁错了。

现在问最少的路线长度。

这个路线是每次选择往左走还是往右走，因为每个点有两条边

E Entirely Unsorted Sequences

题意：

有\(n\)个数，定义一个数是有序的当且仅当它左边的数都小于等于它，它右边的数都大于等于它。

现在问有多少种这些数的排列，使得所有数都不是有序的。

F Financial Planning

题意：

给出一些理财产品，对于每个理财产品，刚开始需要投入\(c_i\)的钱，之后每一天都会获得\(p_i\)的钱。

过了\(x\)天，一款理财产品带来的收益是\(xp_i - c_i\)。

现在问，如果选择理财产品，使得\(x\)天后的收益大于等于\(M\)

要满足\(x\)最小

思路：

二分\(x\)，那么每款理财产品的收益固定，贪心的取。

view code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
int n, m, p[N], c[N];
bool check(ll x) {
	ll res = 0;
	for (int i = 1; i <= n; ++i) {
		ll t = 1ll * p[i] * x - c[i];
		res += max(0ll, t);
		if (res >= m) return true;
	}
	return res >= m;
}

int main() {
	while (scanf("%d%d", &n, &m) != EOF) {
		for (int i = 1; i <= n; ++i) scanf("%d%d", p + i, c + i);
		ll l = 0, r = 2e9, res = 2e9;
		while (r - l >= 0) {
			ll mid = (l + r) >> 1;
			if (check(mid)) {
				r = mid - 1;
				res = mid;
			} else {
				l = mid + 1;
			}
		}
		printf("%lld\n", res);
	}
	return 0;
}

G Game Night

题意：

有一个长度为\(n\)的字符串环，里面只有'A', 'B', 'C'三种字符。

问最少要移动多少个字符，使得同类字符所在的位置的连续的。

思路：

枚举最终形态，那么如果一个位置最终形态对应的位置不是本身，那么这个字符要动。

维护三个前缀和\(O(1)\)算贡献。

view code

#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 10;
int n, A[N], B[N], C[N]; char s[N];
int get(char c, int l, int r) {
	if (l > r) return 0;
	if (c == 'A') return (A[r] - A[l - 1]);
	else if (c == 'B') return (B[r] - B[l - 1]);
	return (C[r] - C[l - 1]);
}
int gao(string t) {
	int num[3];
	for (int i = 0; i < 3; ++i)
		num[i] = get(t[i], 1, n);
	int res = 0;
	for (int i = 0; i <= num[0]; ++i) {
		res = max(res, get(t[0], 1, i) + get(t[1], i + 1, i + num[1]) + get(t[2], i + num[1] + 1, i + num[1] + num[2]) + get(t[0], i + num[1] + num[2] + 1, n));
	}
	return n - res;
}

int main() {
	while (scanf("%d%s", &n, s + 1) != EOF) {
		memset(A, 0, sizeof A);
		memset(B, 0, sizeof B);
		memset(C, 0, sizeof C);
		for (int i = 1; i <= n; ++i) {
			A[i] = A[i - 1] + (s[i] == 'A');
			B[i] = B[i - 1] + (s[i] == 'B');
			C[i] = C[i - 1] + (s[i] == 'C');
		}
		int res = 1e9;
		string t = "ABC";
		do {
			res = min(res, gao(t));
		} while (next_permutation(t.begin(), t.end()));
		printf("%d\n", res);
	}
	return 0;
}

H Harry the Hamster

题意：

有一张\(n\)个点\(m\)条边的有向图，每条边有边权，一个人在\(S\)点，它的左脑不想睡觉，它的右脑想睡觉。

两个脑袋轮流操作，每次操作在当前点选择一条出边往下走。

保证除了\(T\)点之外每个点都有出边，并且\(T\)没有出边。

问两个脑袋都最优操作，最终能否到\(T\)，能的话就输出到\(T\)的时间，不能的话就输出'infinity'

I In Case of an Invasion, Please. . .

题意：

给出一张\(n\)个点\(m\)条边的无向图，每个点有\(p_i\)个人，有\(s(1 \leq s \leq 10)\)个保护区。

每个保护区有一个容量\(c_i\)，保证\(\sum c_i \geq \sum p_i\).

现在所有人都要到保护区，令所有人到保护区的最大时间最小。

思路：

二分答案，那么对于每个点，我们可以知道在这个答案下，这个点上的人可以到达哪些保护区。

那么用一个\(s\)的二进制状态表示它的可达状态，发现最多只有\(2^s\)种状态。

将同种状态的点都缩起来，网络流判断是否流量等于\(\sum p_i\)即可。

view code

#include <bits/stdc++.h>

using namespace std;

using ll = long long;

const int N = 2e5 + 10;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;

struct Dicnic {
	static const int M = 2e6 + 10;
	static const int N = 1e4 + 10;

	struct Edge {
		int to, nxt;
		ll flow;

		Edge() {}

		Edge(int to, int nxt, ll flow): to(to), nxt(nxt), flow(flow) {}
	}edge[M];

	int S, T;
	int head[N], tot;
	int dep[N];

	void Init() {
		memset(head, -1, sizeof head);
		tot = 0;
	}

	void set(int _S, int _T) {
		S = _S;
		T = _T;
	}

	void addedge(int u, int v, ll w, ll rw = 0) {
		edge[tot] = Edge(v, head[u], w); head[u] = tot++;
		edge[tot] = Edge(u, head[v], rw); head[v] = tot++;
	}

	bool BFS() {
		memset(dep, -1, sizeof dep);
		queue<int>q;
		q.push(S);
		dep[S] = 1;
		while (!q.empty()) {
			int u = q.front();
			q.pop();
			for (int i = head[u]; ~i; i = edge[i].nxt) {
				if (edge[i].flow && dep[edge[i].to] == -1) {
					dep[edge[i].to] = dep[u] + 1;
					q.push(edge[i].to);
				}
			}
		}
		return dep[T] >= 0;
	}

	ll DFS(int u, ll f) {
		if (u == T || f == 0) {
			return f;
		}
		ll w, used = 0;
		for (int i = head[u]; ~i; i = edge[i].nxt) {
			if (edge[i].flow && dep[edge[i].to] == dep[u] + 1) {
				w = DFS(edge[i].to, min(f - used, edge[i].flow));
				edge[i].flow -= w;
				edge[i ^ 1].flow += w;
				used += w;
				if (used == f) return f;
			}
		}
		if (!used) dep[u] = -1;
		return used;
	}

	ll solve() {
		ll res = 0;
		while (BFS()) {
			res += DFS(S, INFLL);
		}
		return res;
	}
}dicnic;

struct Edge {
	int to, nxt;
	ll w;

	Edge() {}

	Edge(int to, int nxt, ll w): to(to), nxt(nxt), w(w){}
}edge[N << 2];

int n, m, s;
int a[N], c[N], p[N];
int head[N], used[N], tot;
ll dis[11][N];

void Init() {
	memset(head, -1, sizeof head);
	tot = 0;
}

void addedge(int u, int v, int w) {
	edge[tot] = Edge(v, head[u], w); head[u] = tot++;
	edge[tot] = Edge(u, head[v], w); head[v] = tot++;
}

struct qnode {
	int u;
	ll w;

	qnode() {}

	qnode(int u, ll w): u(u), w(w) {}

	bool operator < (const qnode &other) const {
		return w > other.w;
	}
};

void Dij(int id, int S) {
	for (int i = 1; i <= n; ++i) {
		dis[id][i] = INFLL;
		used[i] = false;
	}
	dis[id][S] = 0;
	priority_queue<qnode> pq;
	pq.push(qnode(S, 0));
	while (!pq.empty()) {
		int u = pq.top().u;
		pq.pop();
		if (used[u]) continue;
		used[u] = true;
		for (int i = head[u]; ~i; i = edge[i].nxt) {
			int v = edge[i].to, w = edge[i].w;
			if (!used[v] && dis[id][v] > dis[id][u] + w) {
				dis[id][v] = dis[id][u] + w;
				pq.push(qnode(v, dis[id][v]));
			}
		}
	}
}

ll sum;
ll mark[1 << 11];

bool check(ll x) {
	memset(mark, 0, sizeof mark);
	for (int i = 1; i <= n; ++i) {
		int S = 0;
		for (int j = 1; j <= s; ++j) {
			if (dis[j][i] <= x) {
				S |= 1 << (j - 1);
			}
		}
		mark[S] += p[i];
	}
	dicnic.Init();
	int S = (1 << s) + s + 10, T = S + 1, tmp = 1 << s;
	dicnic.set(S, T);
	for (int i = 0; i < tmp; ++i) {
		dicnic.addedge(S, i, mark[i]);
		for (int j = 0; j < s; ++j) {
			if (i & (1 << j)) {
				dicnic.addedge(i, j + tmp, mark[i]);
			}
		}
	}
	for (int i = (1 << s), j = 1; j <= s; ++j, ++i) {
		dicnic.addedge(i, T, c[j]);
	}
	ll res = dicnic.solve();
	return res == sum;
}

int main() {
	while (scanf("%d %d %d", &n, &m, &s) != EOF) {
		Init();
		sum = 0;
		for (int i = 1; i <= n; ++i) {
			scanf("%d", p + i);
			sum += p[i];
		}
		for (int i = 1, u, v, w; i <= m; ++i) {
			scanf("%d %d %d", &u, &v, &w);
			addedge(u, v, w);
		}
		for (int i = 1; i <= s; ++i) {
			scanf("%d %d", a + i, c + i);
		}
		for (int i = 1; i <= s; ++i) {
			Dij(i, a[i]);
		}
		ll l = 0, r = INFLL, res = INFLL;
		while (r - l >= 0) {
			ll mid = (l + r) >> 1;
			if (check(mid)) {
				r = mid - 1;
				res = mid;
			} else {
				l = mid + 1;
			}
		}
		printf("%lld\n", res);
	}
	return 0;
}

J Janitor Troubles

题意：

给出四条边，问这四条边构成的四边形的最大面积

思路：

四边形是内接圆四边形时最大。

view code

#include <bits/stdc++.h>

using namespace std;

using db = double;

const db eps = 1e-8;

int sgn(db x) {
	if (fabs(x) < eps) return 0;
	else return x > 0 ? 1 : -1;
}

inline db F(db a, db b, db c) {
	db p = (a + b + c) / 2.0;
	db res = sqrt(p * (p - a) * (p - b) * (p - c));
	return res;
}

inline db f(db a, db b, db c, db d) {
	db res = 0.0;
	db l = max(fabs(a - b), fabs(c - d));
	db r = min(fabs(a + b), fabs(c + d));
	for (db e = l; e < r; e += 0.0005) {
		if (a + b - e <= 0) continue;
		if (a + e - b <= 0) continue;
		if (b + e - a <= 0) continue;

		if (c + d - e <= 0) continue;
		if (c + e - d <= 0) continue;
		if (d + e - a <= 0) continue;

		res = max(res, F(a, b, e) + F(c, d, e));
	}
	return res;
}

int a, b, c, d;

int main() {
	while (scanf("%d %d %d %d", &a, &b, &c, &d) != EOF) {
		if (a == b && b == c && c == d) {
			db res = a * b;
			printf("%.10f\n", res);
			continue;
		}
		db res = f(a, b, c, d);
		res = max(res, f(a, c, b, d));
		res = max(res, f(a, d, b, c));
		printf("%.15f\n", res);
	}
	return 0;
}

K Kingpin Escape

题意：

给出一棵树，有一个特殊点，现在可以额外加一些边，使得不管删去原树中的哪条边，每个点仍然可以到达特殊点

view code

#include <bits/stdc++.h>

using namespace std;

int n, m;
vector<vector<int> >G;
vector<int> vec;

void gao(int u, int fa) {
	if (G[u].size() == 1) {
		vec.push_back(u);
		return ;
	}
	for (auto &v : G[u]) if (v != fa){
		gao(v, u);
	}
}

int main() {
	while (scanf("%d %d", &n, &m) != EOF) {
		G.clear();
		G.resize(n);
		vec.clear();
		for (int i = 1, u, v; i < n; ++i) {
			scanf("%d %d", &u, &v);
			G[u].push_back(v);
			G[v].push_back(u);
		}
		if (n == 2) {
			puts("1");
			puts("0 1");
			continue;
		}
		int rt = 0;
		for (int i = 0; i < n; ++i) {
			if (G[i].size() > 1) {
				rt = i;
				break;
			}
		}
		gao(rt, rt);
		int sze = vec.size();
		int cnt = (sze + 1) / 2;
		printf("%d\n", cnt);
		for (int i = 0; i + cnt < sze; ++i) {
			printf("%d %d\n", vec[i], vec[i + cnt]);
		}
		if (sze & 1) {
			if (vec[cnt - 1] != m) {
				printf("%d %d\n", vec[cnt - 1], m);
			} else {
				printf("%d %d\n", vec[cnt - 1], vec[cnt - 2]);
			}
		}
	}
	return 0;
}