UNIQUE VISION Programming Contest 2023 Christmas (AtCoder Beginner Contest 334)
UNIQUE VISION Programming Contest 2023 Christmas (AtCoder Beginner Contest 334)
A - Christmas Present
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 310;
typedef pair<ll, ll> pii;
typedef pair<pii, ll> piii;
#define fi first
#define se second
const int mod = 1e9 + 7;
void solve()
{
int a, b;
cin >> a >> b;
if (a > b)
{
puts("Bat");
}
else
{
puts("Glove");
}
}
int main()
{
int t = 1;
while (t--)
{
solve();
}
return 0;
}
B - Christmas Trees
解题思路:
三种情况：
- \(l \leq a \leq r\)
- \(a \leq l \leq r\)
- \(l \leq r \leq a\)
注意点:\((0,3,6,9)\):\(\frac{6 - 0}{3} = 2\),\(\frac{9 - 0}{3} = 3\),\(3 - 2 = 1\)，但答案是\(2\)。
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 310;
typedef pair<ll, ll> pii;
typedef pair<pii, ll> piii;
#define fi first
#define se second
const int mod = 1e9 + 7;
void solve()
{
ll a, m, l, r;
cin >> a >> m >> l >> r;
ll ans = 0;
if (l <= a && r >= a)
{
ans += (a - l) / m + (r - a) / m + 1;
}
else if (l >= a && r >= a)
{
ans = (r - a) / m - ((l - 1) - a) / m + (l == a ? 1 : 0);
}
else if (l <= a && r <= a)
{
ans = (a - l) / m - (a - (r + 1)) / m + (r == a ? 1 : 0);
}
cout << ans << endl;
}
int main()
{
int t = 1;
while (t--)
{
solve();
}
return 0;
}
C - Socks 2
解题思路:
贪心地想，位置靠的近的优先配对。
如果单只的袜子的个数是偶数个，那么直接升序两两配对。
如果单只的袜子的个数是奇数个，那么必定有一个是无法配对的，我们枚举这个无法配对的单只袜子。
我们发现，如果我们选取第偶数只袜子作为剩下的袜子，一定有一种选取第奇数只袜子作为剩下袜子的选法由于它，所以直接前缀后缀和枚举即可。
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 310;
typedef pair<ll, ll> pii;
typedef pair<pii, ll> piii;
#define fi first
#define se second
const int mod = 1e9 + 7;
void solve()
{
int n, k;
cin >> n >> k;
vector<int> a(1);
for (int i = 1; i <= k; i++)
{
int x;
cin >> x;
a.push_back(x);
}
sort(a.begin() + 1, a.end());
vector<ll> pre(n + 10, 0), suf(n + 10, 0);
n = k;
for (int i = 1; i <= n; i++)
{
if (i & 1)
{
pre[i] = pre[i - 1];
}
else
{
pre[i] = pre[i - 1] + (a[i] - a[i - 1]);
}
}
if (!(n & 1))
{
cout << pre[n] << endl;
return;
}
for (int i = n; i; i--)
{
if (i & 1)
{
suf[i] = suf[i + 1];
}
else
{
suf[i] = suf[i + 1] + (a[i + 1] - a[i]);
}
}
ll ans = 1e18;
for (int i = 1; i <= n; i++)
{
if (i & 1)
{
ans = min(ans, pre[i] + suf[i]);
}
}
cout << ans << endl;
}
int main()
{
int t = 1;
while (t--)
{
solve();
}
return 0;
}
D - Reindeer and Sleigh
解题思路:
预处理升序前缀和，对于每个询问直接二分查找匹配答案。
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 310;
typedef pair<ll, ll> pii;
typedef pair<pii, ll> piii;
#define fi first
#define se second
const int mod = 1e9 + 7;
void solve()
{
int n, q;
cin >> n >> q;
vector<ll> r(n + 1);
for (int i = 1; i <= n; i++)
{
cin >> r[i];
}
sort(r.begin() + 1, r.end());
vector<ll> s(n + 1);
for (int i = 1; i <= n; i++)
{
s[i] = s[i - 1] + r[i];
}
while (q--)
{
ll x;
cin >> x;
auto it = upper_bound(s.begin() + 1, s.end(), x);
it--;
cout << (it - s.begin()) << endl;
}
}
int main()
{
int t = 1;
while (t--)
{
solve();
}
return 0;
}
E - Christmas Color Grid 1
解题思路:
题目说明很清楚，枚举所有的红色位置，用概率乘以修改后全局的连通块数累加即可。
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 310;
typedef pair<ll, ll> pii;
typedef pair<pii, ll> piii;
#define fi first
#define se second
#define int long long
const int mod = 998244353;
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
ll qmi(ll a, ll b)
{
ll res = 1;
while (b)
{
if (b & 1)
{
res = (res * a) % mod;
}
b >>= 1;
a = (a * a) % mod;
}
return res;
}
void solve()
{
int n, m;
cin >> n >> m;
vector<string> g(n + 1);
for (int i = 1; i <= n; i++)
{
cin >> g[i];
g[i] = ' ' + g[i];
}
ll cnt = 0;
vector<vector<int>> id(n + 1, vector<int>(m + 1));
vector<vector<bool>> st(n + 1, vector<bool>(m + 1));
ll uid = 0;
auto check = [&](int a, int b)
{
if (a < 1 || b < 1 || a > n || b > m)
{
return false;
}
if (st[a][b])
{
return false;
}
if (g[a][b] == '.')
{
return false;
}
return true;
};
auto dfs = [&](auto self, int x, int y, int cid) -> void
{
id[x][y] = cid;
st[x][y] = true;
for (int i = 0; i < 4; i++)
{
int nx = x + dx[i];
int ny = y + dy[i];
if (check(nx, ny))
{
self(self, nx, ny, cid);
}
}
};
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
if (!st[i][j] && g[i][j] == '#')
{
uid++;
dfs(dfs, i, j, uid);
}
if (g[i][j] == '.')
{
cnt++;
}
}
}
ll ans = 0;
cnt = qmi(cnt, mod - 2) % mod;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
if (g[i][j] == '.')
{
map<int, bool> mp;
for (int k = 0; k < 4; k++)
{
int a = i + dx[k];
int b = j + dy[k];
// cout << a << ' ' << b << endl;
if (a >= 1 && b >= 1 && a <= n && b <= m && g[a][b] == '#')
{
// cout << id[a][b] << endl;
if (!mp.count(id[a][b]))
{
mp[id[a][b]] = true;
}
}
}
ll t = uid - mp.size() + 1;
ans = ((ans + (t * cnt) % mod + mod) % mod + mod) % mod;
}
}
}
cout << ans << endl;
}
signed main()
{
int t = 1;
while (t--)
{
solve();
}
return 0;
}
F - Christmas Present 2
解题思路:
\(dp[i]:考虑前i个位置的最小总距离。\)
\(dist[i]:从礼物屋子到第i个房子的直线距离\)。
\(d[i][j]:从第i个屋子到大第j个屋子的直线距离\)。
\(pre[i]:从第1个屋子到第i个屋子的顺序总距离\)。
\[\begin{align*}
dp[i] &= min\{dp[j] + dist[j + 1] + \sum_{k = j + 1}^{i - 1}d[k][k + 1] + dist[i]\} \\
&= min\{dp[j] + dist[j + 1] + (pre[i] - pre[j + 1]) + dist[i]\} \\\\
&= dist[i] + pre[i] + min\{(dp[j] + dist[j + 1] - pre[j + 1])\}
\end{align*}
\]
我们设定：
\[f[j] = dp[j] + dist[j + 1] - pre[j + 1]
\]
则
\[dp[i] =dist[i] + pre[i] + min\{f[j]\}
\]
对于上述式子有
\[j \in [max(0,i - k) ,i - 1]
\]
\[dp[i] = 在j点结束后 +从起点到(j + 1)点+从(j+1)顺序走到i点 + 从i点回到礼物屋
\]
枚举所有范围内的\(j\)得到最小值就是\(dp[i]\)的最终答案。
动态维护长度为\(k\)的区间中的最小值，我们可用滑动窗口\(O(n)\)完成。
当然，本题用其他动态查询区间最值的数据结构也可在\(O(nlogn)\)通过。
代码:
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 310;
typedef pair<ll, ll> pii;
typedef pair<pii, ll> piii;
#define fi first
#define se second
const int mod = 998244353;
void solve()
{
int n, k;
cin >> n >> k;
vector<pii> p(n + 1);
for (int i = 0; i <= n; i++)
{
cin >> p[i].fi >> p[i].se;
}
vector<double> pre(n + 1), dist(n + 1);
auto getd = [&](ll x1, ll y1, ll x2, ll y2) -> double
{
return sqrtl((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
};
for (int i = 1; i <= n; i++)
{
dist[i] = getd(p[0].fi, p[0].se, p[i].fi, p[i].se);
if (i > 1)
{
pre[i] = getd(p[i].fi, p[i].se, p[i - 1].fi, p[i - 1].se);
}
}
for (int i = 2; i <= n; i++)
{
pre[i] += pre[i - 1];
}
vector<double> dp(n + 1), f(n + 1);
deque<int> q;
q.push_back(0);
f[0] = dp[0] + dist[1] - pre[1];
// cout << dp[0] << ' ' << dist[1] << endl;
for (int i = 1; i <= n; i++)
{
if (q.size() && q.front() < i - k)
{
q.pop_front();
}
dp[i] = pre[i] + dist[i] + f[q.front()];
f[i] = dp[i] + dist[i + 1] - pre[i + 1];
// cout << i << ' ' << q.front() << endl;
while (q.size() && f[q.back()] >= f[i])
{
q.pop_back();
}
q.push_back(i);
}
// cout << pre[3] << endl;
printf("%.20lf", dp[n]);
}
int main()
{
int t = 1;
while (t--)
{
solve();
}
return 0;
}
G - Christmas Color Grid 2
解题思路:
求割点。
如果将孤立点的绿色变为红色，那么连通块数量减一。
如果将割点处的绿色变为红色，那么连通块数量将增加。增加数取决于周围有多少个邻接连通块。
对于深搜的根节点割点造成的影响要特判一下，与其他正常搜出来的增加数相比要减一。
代码:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<ll, ll>;
#define fi first
#define se second
const int N = 1e6 + 10;
const int mod = 998244353;
vector<string> g(1010);
int n, m;
int cnt;
int dfn[N];
int low[N];
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
int root = 0;
bool cut[1010][1010];
int ch[1010][1010];
bool gu[1010][1010];
ll qmi(ll a, ll b)
{
ll res = 1;
while (b)
{
if (b & 1)
{
res = res * a % mod;
}
a = a * a % mod;
b >>= 1;
}
return res;
}
bool check(int a, int b)
{
if (a < 1 || a > n || b < 1 || b > m)
{
return false;
}
if (g[a][b] == '.')
{
return false;
}
return true;
}
inline void tarjan(int u)
{
dfn[u] = low[u] = ++cnt;
int x = u / m + 1;
int y = u % m;
if (!y)
{
x--;
y = m;
}
int s = 0;
// cout << x << ' ' << y << endl;
for (int i = 0; i < 4; i++)
{
int a = x + dx[i];
int b = y + dy[i];
if (check(a, b))
{
int v = b + ((a - 1) * m);
if (!dfn[v])
{
tarjan(v);
low[u] = min(low[u], low[v]);
if (low[v] >= dfn[u])
{
ch[x][y]++;
// cout << x << ' ' << y << ' ' << ch[x][y] << endl;
if (u != root || ch[x][y] > 1)
{
cut[x][y] = true;
}
}
}
else
{
low[u] = min(low[u], dfn[v]);
}
}
else
{
s++;
}
}
if (s == 4)
{
gu[x][y] = true;
}
}
void solve()
{
cin >> n >> m;
for (int i = 1; i <= n; i++)
{
cin >> g[i];
g[i] = ' ' + g[i];
}
ll tot = 0;
ll num = 0;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
if (g[i][j] == '#')
{
num++;
if (!dfn[j + ((i - 1) * m)])
{
tot++;
root = j + (i - 1) * m;
tarjan(j + ((i - 1) * m));
ch[i][j]--;
}
}
}
}
// cout << tot << endl;
ll t = qmi(num, mod - 2);
// cout << t << endl;
ll ans = 0;
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
if (g[i][j] == '#')
{
if (cut[i][j])
{
// cout << i << ' ' << j << ' ' << ch[i][j] + tot << endl;
ans = (ans + (tot + ch[i][j]) * t) % mod;
}
else if (gu[i][j])
{
ans = (ans + (tot - 1) * t) % mod;
}
else
{
ans = (ans + (tot)*t) % mod;
}
}
}
}
cout << ans << endl;
}
int main()
{
int t = 1;
// cin >> t;
while (t--)
{
solve();
}
return 0;
}
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