AtCoder Beginner Contest 404
A - Not Found
题意
思路
模拟
代码
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#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 1e6 + 10;
void solve()
{
string s;
cin >> s;
for (int i = 'a'; i <= 'z'; i++)
{
if (s.find(i) == s.npos)
{
cout << (char)i << endl;
return;
}
}
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
//cin >> __;
while (__--) solve();
return 0;
}
B - Grid Rotation header
题意
思路
不旋转答案就是二者不同元素的个数，再加上旋转次数即可
代码
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#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 1e6 + 10;
int n;
vector<string> a, b;
void f()
{
vector<string> t(n, string(n, '.'));
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
t[j][n - 1 - i] = a[i][j];
}
}
a = t;
}
void solve()
{
cin >> n;
a.resize(n);
b.resize(n);
for (int i = 0; i < n; i++)
{
cin >> a[i];
}
for (int i = 0; i < n; i++)
{
cin >> b[i];
}
int ans = 1e18;
for (int i = 0; i < 4; i++)
{
int dif = 0;
for (int i = 0; i < n; i++)
{
for (int j = 0; j < n; j++)
{
if (a[i][j] != b[i][j]) dif++;
}
}
ans = min(ans, i + dif);
f();
}
cout << ans << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
//cin >> __;
while (__--) solve();
return 0;
}
C - Cycle Graph?
题意
思路
顶点度数都为\(2\)且连通的简单无向图即循环图
代码
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#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 1e6 + 10;
void solve()
{
int n, m;
cin >> n >> m;
vector<int> deg(n, 0);
vector<vector<int>> g(n);
for (int i = 0; i < m; i++)
{
int u, v;
cin >> u >> v;
u--, v--;
g[u].push_back(v);
g[v].push_back(u);
deg[u]++;
deg[v]++;
}
if (n != m)
{
cout << "No" << endl;
return;
}
for (auto& i : deg)
{
if (i != 2)
{
cout << "No" << endl;
return;
}
}
vector<bool> vis(n, false);
queue<int> q;
q.push(0);
vis[0] = true;
int cnt = 1;
while (!q.empty())
{
int u = q.front();
q.pop();
for (int v : g[u])
{
if (!vis[v])
{
vis[v] = true;
q.push(v);
cnt++;
}
}
}
cout << (cnt == n ? "Yes" : "No") << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
//cin >> __;
while (__--) solve();
return 0;
}
D - Goin' to the Zoo
题意
思路
每种动物只需要看\(2\)次，则一个动物园的访问次数最多也为\(2\)，暴力枚举所有情况
代码
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#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 1e6 + 10;
void solve()
{
int n, m;
cin >> n >> m;
vector<int> c(n);
for (int i = 0; i < n; i++)
{
cin >> c[i];
}
vector<vector<int>> a(n);
for (int i = 0; i < m; i++)
{
int k;
cin >> k;
for (int j = 0; j < k; j++)
{
int x;
cin >> x;
x--;
a[x].push_back(i);
}
}
vector<int> p3(n + 1, 1);
for (int i = 1; i <= n; i++)
{
p3[i] = p3[i - 1] * 3;
}
int ans = LLONG_MAX;
for (int i = 0; i < p3[n]; i++)
{
vector<int> cnt(m, 0);
int res = 0;
for (int j = 0; j < n; j++)
{
int t = i / p3[j] % 3; // 该动物园的访问次数
while (t--)
{
for (auto& l : a[j]) cnt[l]++;
res += c[j];
}
}
if (*min_element(cnt.begin(), cnt.end()) >= 2) ans = min(ans, res);
}
cout << ans << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
//cin >> __;
while (__--) solve();
return 0;
}
E - Bowls and Beans
题意
思路
\(f_i = min\{f_{i-c_i},f_{i-c_i+1},···,f_{i-1}\}+1\)
代码
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#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
typedef pair<int, int> pii;
const int mxn = 1e6 + 10;
void solve()
{
int n, ans = 0;
cin >> n;
vector<int> a(n), c(n), f(n, 0);
a[0] = c[0] = 0;
for (int i = 1; i < n; i++)
{
cin >> c[i];
}
for (int i = 1; i < n; i++)
{
cin >> a[i];
}
for (int i = 1; i < n; i++)
{
f[i] = 1e18;
for (int j = i - c[i]; j < i; j++)
{
f[i] = min(f[i], f[j] + 1);
}
if (a[i])
{
ans += f[i];
// 从其他碗转移来的豆子都可以一次性拿走
f[i] = 0;
}
}
cout << ans << endl;
}
signed main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int __ = 1;
//cin >> __;
while (__--) solve();
return 0;
}
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