[ABC350] 賽後總結
AK 之。
A
模擬
// Problem: A - Past ABCs
// Contest: AtCoder - AtCoder Beginner Contest 350
// Author: Moyou
// Copyright (c) 2024 Moyou All rights reserved.
// Date: 2024-04-20 20:00:23
#include <algorithm>
#include <cstring>
#include <iostream>
#include <queue>
using namespace std;
const int N = 2e5 + 10;
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
string s;
cin >> s;
int x = atoi(s.substr(3, 3).c_str());
if(x >= 1 && x <= 349 && x != 316) cout << "Yes\n";
else cout << "No\n";
return 0;
}
B
模擬
C
貪心，每次和當前位置下標交換，維護對映。
// Problem: C - Sort
// Contest: AtCoder - AtCoder Beginner Contest 350
// Author: Moyou
// Copyright (c) 2024 Moyou All rights reserved.
// Date: 2024-04-20 20:03:21
#include <algorithm>
#include <cstring>
#include <iostream>
#include <queue>
// #define int long long
using namespace std;
const int N = 2e5 + 10;
int n, a[N], b[N];
vector<pair<int, int> > ans;
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> n;
for(int i = 1; i <= n; i ++) cin >> a[i], b[a[i]] = i;
for(int i = 1; i < n; i ++) {
if(a[i] != i) {
ans.push_back({i, b[i]});
int x = i, y = b[i], t = a[i];
swap(a[i], a[b[i]]);
b[a[x]] = x, b[a[y]] = y;
}
}
cout << ans.size() << '\n';
for(auto [x, y] : ans) cout << x << ' ' << y << '\n';
return 0;
}
D
連通塊內完全圖。
// Problem: D - New Friends
// Contest: AtCoder - AtCoder Beginner Contest 350
// Author: Moyou
// Copyright (c) 2024 Moyou All rights reserved.
// Date: 2024-04-20 20:22:21
#include <algorithm>
#include <cstring>
#include <iostream>
#include <queue>
#define int long long
using namespace std;
const int N = 2e5 + 10;
int n, m, a[N], b[N];
int fa[N], rk[N], top;
int find(int x) {return fa[x] == x ? x : fa[x] = find(fa[x]); }
void merge(int a, int b) {
int x = find(a), y = find(b);
if(x == y) return ;
if(rk[x] > rk[y]) swap(x, y);
fa[x] = y, rk[y] += rk[x];
}
int ans;
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> n >> m;
for(int i = 1; i <= n; i ++)
fa[i] = i, rk[i] = 1;
for(int i = 1, a, b; i <= m; i ++) {
cin >> a >> b;
merge(a, b);
}
for(int i = 1; i <= n; i ++)
if(fa[i] == i) ans += 1ll * rk[i] * (rk[i] - 1) / 2;
cout << ans - m << '\n';
return 0;
}
E
直接記憶化搜尋即可，狀態不多。
// Problem: E - Toward 0
// Contest: AtCoder - AtCoder Beginner Contest 350
// Author: Moyou
// Copyright (c) 2024 Moyou All rights reserved.
// Date: 2024-04-20 20:27:56
#include <algorithm>
#include <map>
#include <iostream>
#include <queue>
#define int long long
using namespace std;
const int N = 1e7 + 10;
map<int, double> f;
int n, a, x, y;
double DP(int n) {
if(n == 0) return 0;
if(f.count(n)) return f[n];
double sum = 0;
for(int i = 2; i <= 6; i ++)
sum += DP(n / i);
return f[n] = min(DP(n / a) + x, 1.0 * (sum + 6 * y) / 5);
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> n >> a >> x >> y;
printf("%.9lf\n", DP(n));
return 0;
}
F
【模板】文藝平衡樹
// Problem: F - Transpose
// Contest: AtCoder - AtCoder Beginner Contest 350
// Author: Moyou
// Copyright (c) 2024 Moyou All rights reserved.
// Date: 2024-04-20 20:35:01
#include <algorithm>
#include <cstring>
#include <iostream>
#include <queue>
#include <ctime>
#include <random>
// #define int long long
using namespace std;
const int N = 5e5 + 10;
int stk[N], top;
char ope(char ch) {
if(islower(ch)) return toupper(ch);
return tolower(ch);
}
struct qwq {
int l, r;
char key;
int val, size;
bool tag;
} tr[N];
int root, idx;
mt19937 rando(time(0));
#define l(k) tr[k].l
#define r(k) tr[k].r
int New(char val) {return tr[++ idx] = {0, 0, val, rando(), 1}, idx; }
void up(int k) {tr[k].size = tr[l(k)].size + tr[r(k)].size + 1; }
void down(int k)
{
if(!tr[k].tag) return ;
if(tr[k].key != '#') tr[k].key = ope(tr[k].key);
swap(l(k), r(k));
if(l(k)) tr[l(k)].tag ^= 1;
if(r(k)) tr[r(k)].tag ^= 1;
tr[k].tag = 0;
}
void split(int u, int x, int &a, int &b)
{
if(!u) {a = b = 0; return ;}
down(u);
if(tr[l(u)].size + 1 <= x) split(r(u), x - tr[l(u)].size - 1, r(u), b), a = u;
else split(l(u), x, a, l(u)), b = u;
up(u);
}
int merge(int a, int b)
{
if(!a || !b) return a + b;
if(tr[a].val > tr[b].val) {
down(a), r(a) = merge(r(a), b), up(a);
return a;
}
else {
down(b), l(b) = merge(a, l(b)), up(b);
return b;
}
}
void rev(int l, int r) {
int a, m, b;
split(root, r, a, b);
split(a, l - 1, a, m);
tr[m].tag ^= 1;
root = merge(a, merge(m, b));
}
void print(int u) {
if(!u) return ;
bool f = tr[u].tag;
down(u);
print(l(u));
if(tr[u].key != '#') {
cout << tr[u].key;
}
print(r(u));
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
string s;
cin >> s;
int n = s.size();
s = " " + s;
for(int i = 1; i <= n; i ++) {
if(isalnum(s[i])) {
root = merge(root, New(s[i]));
}
else root = merge(root, New('#'));
}
for(int i = 1; i <= n; i ++) {
if(s[i] == '(') stk[++ top] = i;
else if(s[i] == ')') {
rev(stk[top], i);
top --;
}
}
print(root);
return 0;
}
也可以利用括號匹配的性質遞迴做，但是文藝無腦好調快。
G
按度數根號分治，對於兩個點度數都很大的情況需要支援 __builtin_clz 函式，但是我不知道，所以這一部分暴力過掉了。
// Problem: G - Mediator
// Contest: AtCoder - AtCoder Beginner Contest 350
// Author: Moyou
// Copyright (c) 2024 Moyou All rights reserved.
// Date: 2024-04-20 20:56:32
#include <algorithm>
#include <cstring>
#include <iostream>
#include <queue>
#include <bitset>
// #define int long long
using namespace std;
const int N = 1e5 + 10, B = 320, mod = 998244353;
int n, q, fa[N];
bool ok[N], is[B][B];
bitset<N> d[B];
vector<int> g[N];
int h[N], idx;
int id(int x) {
if(h[x]) return h[x];
return h[x] = ++ idx;
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> n >> q;
for(int i = 1, op, a, b, lst = 0; i <= q; i ++) {
cin >> op >> a >> b;
op = 1 + ((1ll * op * (1 + lst)) % mod) % 2;
a = 1 + ((1ll * a * (1 + lst)) % mod) % n;
b = 1 + ((1ll * b * (1 + lst)) % mod) % n;
bool x = g[a].size() > B, y = g[b].size() > B;
if(op == 1) {
g[a].push_back(b), g[b].push_back(a);
if(!x && g[a].size() > B) {
for(auto v : g[a]) {
d[id(a)][v] = 1;
if(g[v].size() > B) is[id(v)][id(a)] = is[id(a)][id(v)] = 1;
}
}
if(!y && g[b].size() > B) {
for(auto v : g[b]) {
d[id(b)][v] = 1;
if(g[v].size() > B) is[id(v)][id(b)] = is[id(b)][id(v)] = 1;
}
}
if(g[a].size() > B) d[id(a)][b] = 1;
if(g[b].size() > B) d[id(b)][a] = 1;
if(g[a].size() > B && g[b].size() > B) is[id(a)][id(b)] = is[id(b)][id(a)] = 1;
}
else {
if(x && y) {
if(is[id(a)][id(b)]) {
cout << (lst = 0) << '\n';
continue;
}
auto t = d[id(a)] & d[id(b)];
if(t.none()) cout << (lst = 0) << '\n';
else {
for(int i = 1; i <= n; i ++) {
if(t[i]) {
lst = i;
break;
}
}
cout << (lst) << '\n';
}
}
else if(!x && !y) {
for(auto v : g[a]) ok[v] = 1;
lst = 0;
for(auto v : g[b]) if(ok[v]) {
lst = v;
break;
}
cout << lst << '\n';
for(auto v : g[a]) ok[v] = 0;
}
else {
if(x) swap(a, b), swap(x, y);
lst = 0;
for(auto v : g[a])
if(d[id(b)][v]) {
lst = v;
break;
}
cout << lst << '\n';
}
}
}
return 0;
}
啟發式合併也可以做，每次合併的時候重構小連通塊的形態，查詢分討容易做。
時間複雜度：\(O(n\log n)\)。
// Problem: G - Mediator
// Contest: AtCoder - AtCoder Beginner Contest 350
// Author: Moyou
// Copyright (c) 2024 Moyou All rights reserved.
// Date: 2024-04-20 20:56:32
#include <algorithm>
#include <cstring>
#include <iostream>
#include <queue>
using namespace std;
const int N = 1e5 + 10, mod = 998244353;
int n, q, fa[N], pa[N], sz[N];
vector<int> g[N];
bool st[N];
int find(int x) {return pa[x] == x ? x : pa[x] = find(pa[x]);}
void merge(int a, int b) {
int x = find(a), y = find(b);
if(x == y) return ;
pa[x] = y, sz[y] += sz[x];
}
void dfs(int u, int f) {
fa[u] = f;
for(auto v : g[u]) {
if(v == f) continue;
dfs(v, u);
}
}
signed main() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cin >> n >> q;
for(int i = 1; i <= n; i ++) pa[i] = i, sz[i] = 1;
for(int i = 1, op, a, b, lst = 0; i <= q; i ++) {
cin >> op >> a >> b;
op = 1 + ((1ll * op * (1 + lst)) % mod) % 2;
a = 1 + ((1ll * a * (1 + lst)) % mod) % n;
b = 1 + ((1ll * b * (1 + lst)) % mod) % n;
if(op == 1) {
if(sz[find(a)] > sz[find(b)]) swap(a, b);
merge(a, b);
dfs(a, b);
g[a].push_back(b), g[b].push_back(a);
}
else {
if(fa[fa[a]] == b) cout << (lst = fa[a]) << '\n';
else if(fa[fa[b]] == a) cout << (lst = fa[b]) << '\n';
else if(fa[a] == fa[b]) cout << (lst = fa[a]) << '\n';
else cout << (lst = 0) << '\n';
}
}
return 0;
}
總結
C 切太慢了，以為是快排，對於這種對映要細心。
G 大膽可過。