2023ICPC网络赛第一场 - A D G J L
D
榜单：传送门
题解：知乎链接
感觉知乎的题解已经很充分了，这里仅分享AC代码
LADJG
A 模拟
D 图的连通块
G 树
J 期望?
L 模拟
A Qualifiers Ranking Rules
模拟题给的操作流程，关键在于去重
#include<bits/stdc++.h>
#define ll long long
using namespace std;
void solve(){
int n, m;
cin >> n >> m;
string ss;
vector<string> s, t;
map<string, int> qs, qt;
for(int i = 0; i < n; ++ i){
cin >> ss;
if(!qs.count(ss)){
qs[ss] = 1;
s.push_back(ss);
}
}
for(int i = 0; i < m; ++ i){
cin >> ss;
if(!qt.count(ss)){
qt[ss] = 1;
t.push_back(ss);
}
}
int i = 0, j = 0;
map<string, int> qa;
while(i < s.size() || j < t.size()){
if(i < s.size()){
if(!qa.count(s[i])){
cout << s[i] << '\n';
qa[s[i]] = 1;
}
++ i;
}
if(j < t.size()){
if(!qa.count(t[j])){
cout << t[j] << '\n';
qa[t[j]] = 1;
}
++ j;
}
}
return ;
}
signed main(){
int t = 1;
// cin >> t;
while(t --){
solve();
}
return 0;
}
D Transitivity
考虑原无向图构成的所有连通块
如果有连通块不是完全图，将其补成完全图即可，需要补的边数即为 $n * (n - 1) / 2 - m $
如果所有连通块都是完全图，那么我们只需要拿顶点数最少的两个连通块将其点两两联通即可
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll n, m;
struct DSU{
int num;
vector<int> fa, sz, edge;
DSU(int x) : num(x), fa(x + 1), sz(x + 1, 1), edge(x + 1, 0){
for(int i = 0; i <= x; ++ i){
fa[i] = i;
}
}
int size(int x){ return sz[findfa(x)]; }
int edgesize(int x){ return edge[findfa(x)]; }
int findfa(int x){
while(x != fa[x]) x = fa[x] = fa[fa[x]];
return x;
}
bool same(int x, int y){ return findfa(x) == findfa(y); }
bool merge(int x, int y){
x = findfa(x); y = findfa(y);
++ edge[x];
if(x == y) return false;
sz[x] += sz[y];
edge[x] += edge[y];
fa[y] = x;
return true;
}
};
void solve(){
cin >> n >> m;
DSU dsu(n);
for(int i = 0; i < m; ++ i){
int u, v;
cin >> u >> v;
dsu.merge(u, v);
}
ll res = 0, ans = 1;
priority_queue<ll, vector<ll>, greater<ll>> q;
map<int, int> vis;
for(int i = 1; i <= n; ++ i){
int f = dsu.findfa(i);
if(!vis.count(f)){
vis[f] = 1;
ll size = dsu.sz[f];
if(size != 1)
res += size * (size - 1) / 2 - dsu.edgesize(f);
q.push(size);
}
}
int cs = 0;
if(q.size() != 1){
while(cs < 2){
ans *= q.top();
q.pop();
++ cs;
}
}else ans = 0;
if(res) cout << res << '\n';
else cout << ans << '\n';
return ;
}
signed main(){
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int t = 1;
// cin >> t;
while(t --){
solve();
}
return 0;
}
G Spanning Tree
#include<bits/stdc++.h>
#define ll long long
using namespace std;
typedef pair<int, int> pii;
const int maxm = 1e6;
ll mod = 998244353;
ll n;
ll qpow(ll a, ll x){
a %= mod;
ll res = 1;
while(x){
if(x & 1) res = res * a % mod;
a = a * a % mod;
x >>= 1;
}
return res;
}
struct DSU{
int num;
vector<int> fa, sz;
DSU(int x) : num(x), fa(x + 1), sz(x + 1, 1){
for(int i = 0; i <= x; ++ i){
fa[i] = i;
}
}
int size(int x){ return sz[findfa(x)]; }
int findfa(int x){
while(x != fa[x]) x = fa[x] = fa[fa[x]];
return x;
}
bool same(int x, int y){ return findfa(x) == findfa(y); }
bool merge(int x, int y){
x = findfa(x); y = findfa(y);
if(x == y) return false;
sz[x] += sz[y];
fa[y] = x;
return true;
}
};
void solve(){
cin >> n;
vector<pii> s;
for(int i = 0; i < n - 1; ++ i){
int u, v;
cin >> u >> v;
s.push_back({u, v});
}
vector<pii> t;
vector<int> e[n + 1];
for(int i = 0; i < n - 1; ++ i){
int u, v;
cin >> u >> v;
e[u].push_back(v);
e[v].push_back(u);
t.push_back({u, v});
}
DSU dsu(n);
ll ans = 1;
bool f = false;
for(auto &[u, v] : s){
u = dsu.findfa(u);
v = dsu.findfa(v);
ll su = dsu.size(u), sv = dsu.size(v);
ans = ans * su % mod * sv % mod;
if(e[u].size() < e[v].size()) swap(u, v);
int cnt = 0;
for(auto x : e[v]){
if(dsu.findfa(x) == u) ++ cnt;
else e[u].push_back(x);
}
if(cnt != 1){
f = true; break;
}
dsu.merge(u, v);
}
if(f) cout << "0\n";
else{
ans = qpow(ans, mod - 2);
cout << ans << '\n';
}
return ;
}
signed main(){
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int t = 1;
// cin >> t;
while(t --){
solve();
}
return 0;
}
J Minimum Manhattan Distance
题意
在圆 C2 上找一点使得它距离圆 C1 上等概率随机一点的期望曼哈顿距离最短，并求出这个期望
思路
期望是一种加权平均
当随机变量 X 取各个可能值是等概率分布的时候，X 的期望值与算术平均值相等
由于具有对称性，最小期望曼哈顿距离就是该点到 C1 圆心的曼哈顿距离
由曼哈顿距离的等高线可以得知，答案点一定在 C2 四个 $45° $ 方向上，再次观察计算可以发现，答案即为 两圆心的曼哈顿距离减去圆 C2 半径的 $\sqrt{2} $ 倍
代码
#include<bits/stdc++.h>
#define ll long long
using namespace std;
void solve(){
ll a, b, c, d, A, B, C, D;
cin >> a >> b >> c >> d;
cin >> A >> B >> C >> D;
double x, y, xx, yy;
x = 1.0 * (a + c) / 2.0;
y = 1.0 * (b + d) / 2.0;
xx = 1.0 * (A + C) / 2.0;
yy = 1.0 * (B + D) / 2.0;
double r2 = sqrt((C - A) * (C - A) + (D - B) * (D - B)) / 2.0;
double ans = abs(xx - x) + abs(yy - y) - sqrt(2.0) * r2;
cout << fixed << setprecision(15) << ans << '\n';
return ;
}
signed main(){
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
int t = 1;
cin >> t;
while(t --){
solve();
}
return 0;
}
L KaChang!
$O(n) $ 暴力判断每个时间即可
#include<bits/stdc++.h>
#define ll long long
using namespace std;
void solve(){
ll n, t;
cin >> n >> t;
vector<ll> a(n);
for(int i = 0; i < n; ++ i){
cin >> a[i];
}
ll k = 2;
for(int i = 0; i < n; ++ i){
k = max(k, a[i] / t + (a[i] % t != 0));
}
cout << k << '\n';
return ;
}
signed main(){
int t = 1;
// cin >> t;
while(t --){
solve();
}
return 0;
}
