Atcoder Grand Contest 001 题解

最近在刷AGC，就写一下题解。

A - BBQ Easy

sort完，每相邻两个组成一组，发现这样肯定是最优的。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
int n, L[210], ans;
 
int main()
{
	gi(n);
	for (int i = 1; i <= (n << 1); ++i) gi(L[i]);
	sort(L + 1, L + 2 * n + 1);
	for (int i = 1; i <= (n << 1); i += 2) ans += L[i];
	printf("%d\n", ans);
	return 0;
}

　　

B - Mysterious Light

把光线分成很多平行四边形，所有平行四边形做法一样，只要递归下去就好了，只要某一边是另一边倍数就到达终点了，所以复杂度和gcd一样，是log的。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
long long ans = 0;
 
void work(long long x, long long y)
{
	if (x < y)
	{
		ans += (y / x) * x * 2 - x;
		if (y % x == 0)
			return;
		ans += x;
		work(x, y % x);
	}
	else
	{
		ans += (x / y) * y * 2 - y;
		if (x % y == 0)
			return;
		ans += y;
		work(y, x % y);
	}
}
 
int main()
{
	long long N, X;
	cin >> N >> X;
	ans += N;
	work(X, N - X);
	cout << ans << endl;
}

　　

C - Shorten Diameter

枚举根，如果是偶数，所有深度要小于等于k/2，如果是奇数，则有一个根的儿子里所有点的深度可以最大为k/2+1，其他都为k/2。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
const int N = 2010;
 
VI edge[N];
 
int n, k, dep[N], mxd;
 
void dfs(int u, int fa, int d)
{
	if (k & 1) if(d > k / 2 + 1) return;
	if (!(k & 1)) if (d > k / 2) return;
	++dep[d];
	mxd = max(mxd, d);
	for (auto v : edge[u])
		if (v != fa)
		{
			dfs(v, u, d + 1);
		}
}
 
int main()
{
	gii(n, k);
	for (int i = 1; i < n; ++i)
	{
		int u, v;
		gii(u, v);
		edge[u].pb(v);
		edge[v].pb(u);
	}
	int res = 0;
	for (int root = 1; root <= n; ++root)
	{
		int mx = 0, s = 1;
		for (auto son : edge[root])
		{
			mxd = 0;
			dfs(son, root, 1);
			if (k & 1) mx = max(mx, dep[k / 2 + 1]);
			if (k & 1) s -= dep[k / 2 + 1];
			for (int i = 1; i <= mxd; ++i) s += dep[i], dep[i] = 0;
		}
		res = max(res, s + mx);
	}
	printf("%d\n", n - res);
	return 0;
}

　　

D - Arrays and Palindrome

有一个结论：奇数不能超过2，否则无解，有解的话就把奇数放在两端，然后每次错1位放b就能保证他们全部相等了。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
int n, m, a[110];
 
bool comp(const int &i, const int &j)
{
	return (i & 1) > (j & 1);
}
 
int b[110], ans;
 
int main()
{
	gii(n, m);
	int cnt = 0;
	for (int i = 1; i <= m; ++i) gi(a[i]);
	for (int i = 1; i <= m; ++i)
		if (a[i] & 1) ++cnt;
	if (cnt > 2)
	{
		puts("Impossible");
		return 0;
	} 
	sort(a + 1, a + m + 1, comp);
	if (a[2] & 1) swap(a[2], a[m]);
	for (int i = 1; i <= m; ++i)
		printf("%d ", a[i]);
	puts("");
	if (m == 1)
	{
		if (a[1] - 1) printf("2\n%d %d \n", 1, a[1] - 1);
		else printf("1\n%d \n", 1);
		return 0;
	}
	if (a[1] - 1) b[++ans] = a[1] - 1;
	for (int i = 2; i < m; ++i)
		b[++ans] = a[i];
	b[++ans] = a[m] + 1;
	printf("%d\n", ans);
	for (int i = 1; i <= ans; ++i)
		printf("%d ", b[i]);
	puts("");
	return 0;
}

　　

E - BBQ Hard

首先答案是sigma(i<j) C(a[i]+a[j]+b[i]+b[j],a[i]+a[j])，一个C(a[i]+a[j]+b[i]+b[j],a[i]+a[j])可以看作从(-a[i],-b[i])走到(a[j],b[j])的方案数，所以我们所有dp一次性做了，减去一个点自己到自己，最后除以2。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
const int mod = 1e9 + 7;
 
int n;
 
int dp[4010][4010];
 
PII x[200010];
 
int& f(int x, int y) { return dp[x + 2005][y + 2005]; }
 
int inc(int x, int y)
{
	x += y;
	if (x >= mod) x -= mod;
	return x;
}
 
int dec(int x, int y)
{
	x -= y;
	if (x < 0) x += mod;
	return x;
}
 
int fpow(int a, int x)
{
	int ret = 1;
	for (; x; x >>= 1)
	{
		if (x & 1) ret = 1LL * ret * a % mod;
		a = 1LL * a * a % mod;
	}
	return ret;
} 
 
int fac[8010], ifac[8010];
 
int C(int n, int m)
{
	if (n < m || m < 0) return 0;
	return 1LL * fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
 
int main()
{
	fac[0] = 1;
	for (int i = 1; i <= 8000; ++i)
		fac[i] = 1LL * fac[i - 1] * i % mod;
	ifac[8000] = fpow(fac[8000], mod - 2);
	for (int i = 8000; i; --i)
		ifac[i - 1] = 1LL * ifac[i] * i % mod;
	gi(n);
	for (int i = 1; i <= n; ++i)
		gii(x[i].fi, x[i].se), ++f(-x[i].fi, -x[i].se);
	for (int i = -2000; i <= 2000; ++i)
		for (int j = -2000; j <= 2000; ++j)
			f(i, j) = inc(f(i, j), f(i - 1, j)), f(i, j) = inc(f(i, j), f(i, j - 1));
	int ans = 0;
	for (int i = 1; i <= n; ++i)
		ans = inc(ans, f(x[i].fi, x[i].se));
	for (int i = 1; i <= n; ++i)
		ans = dec(ans, C((x[i].fi + x[i].se) << 1, x[i].fi << 1));
	ans = 1LL * ans * fpow(2, mod - 2) % mod;
	printf("%d\n", ans);
	return 0;
}

　　

F - Wide Swap

我们可以令q[p[i]]=i，那么就是相邻的如果|q[i]-q[i+1]|>=k就能交换，所以相对顺序不变，也就是|i-j|<k,p[i]<p[j]是固定的。如果i向j连边就是拓扑排序问题，边是n^2的就线段树优化一下。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
const int N = 5e5 + 10;
 
int n, k, p[N], q[N];
 
int t[N << 2], cnt;
 
int get(int x, int y)
{
	if (!x) return y;
	if (!y) return x;
	return min(x, y);
}
 
void modify(int x, int v)
{
	t[x += cnt] = v;
	for (x >>= 1; x; x >>= 1)
		t[x] = get(t[x << 1], t[x << 1 | 1]);
}
 
int query(int l, int r)
{
	int ans = 0;
	l = max(l, 1);
	r = min(r, n);
	if (l > r) return 0;
	for (l += cnt - 1, r += cnt + 1; l ^ r ^ 1; l >>= 1, r >>= 1)
	{
		if (~l & 1) ans = get(ans, t[l ^ 1]);
		if ( r & 1) ans = get(ans, t[r ^ 1]);
	}
	return ans;
}
 
vector<int> edge[N];
 
int deg[N];
 
priority_queue<int> pq;
 
int ans[N];
 
int main()
{
	gii(n, k);
	for (int i = 1; i <= n; ++i) gi(p[i]), q[p[i]] = i;
	for (cnt = 1; cnt < n + 2; cnt <<= 1); --cnt;
	for (int i = n; i; --i)
	{
		int x = q[query(q[i] - k + 1, q[i] - 1)];
		if (x) edge[q[i]].pb(x), ++deg[x];
		x = q[query(q[i] + 1, q[i] + k - 1)];
		if (x) edge[q[i]].pb(x), ++deg[x];
		modify(q[i], i);
	}
	for (int i = 1; i <= n; ++i)
		if (!deg[i]) pq.push(-i);
	int cur = 0;
	while (!pq.empty())
	{
		int u = -pq.top(); pq.pop();
		ans[u] = ++cur;
		for (auto v : edge[u])
		{
			--deg[v];
			if (!deg[v])
				pq.push(-v);
		}
	}
	for (int i = 1; i <= n; ++i)
		printf("%d\n", ans[i]);
	return 0;
}