The 2024 ICPC Asia East Continent Final Contest
Preface
好久没训练了，一想到现在的水平下周要去打 CCPC Final，感觉冲击 Ag 都成问题啊
这场直接头等战犯，先是开局签到 A 连 WA 四发搞崩罚时，后期模拟 E 怒写半天最后还因为 Corner Case 没过
最后把理论 6 题希望冲 Au 的局打崩了，直接俯冲 Cu 区
A. Hitoshizuku
考虑从小到大考虑每个 \(b_i\) 最小的人，显然让他去匹配两个 \(a_j\le b_i\) 且 \(b_j\) 最小的两个人一定最优
实现的时候最好直接搞几个 set 来维护，我赛时写的双指针有一些奇怪的细节要注意，WA 了好多发
#include<cstdio>
#include<iostream>
#include<set>
#include<array>
#include<vector>
#include<algorithm>
#define RI register int
#define CI const int&
using namespace std;
const int N=300005;
int t,n,vis[N]; array <int,3> a[N],b[N];
int main()
{
for (scanf("%d",&t);t;--t)
{
scanf("%d",&n); n*=3;
for (RI i=1;i<=n;++i)
{
int x,y; scanf("%d%d",&x,&y);
a[i]={x,y,i}; b[i]={y,x,i}; vis[i]=0;
}
sort(a+1,a+n+1); sort(b+1,b+n+1);
bool flag=1; RI pa=1,pb=1,cnt=0;
set <array <int,3>> st;
vector <array <int,3>> ans;
while (cnt<n/3)
{
while (pb<=n&&vis[b[pb][2]]) ++pb;
if (pb>n) { flag=0; break; }
vis[b[pb][2]]=1;
while (pa<=n&&a[pa][0]<=b[pb][0])
{
if (vis[a[pa][2]]) { ++pa; continue; }
st.insert({a[pa][1],a[pa][0],a[pa][2]});
++pa;
}
int num=0,x,y;
while (!st.empty())
{
int id=(*st.begin())[2];
st.erase(st.begin());
if (vis[id]) continue;
++num;
if (num==1) x=id; else
if (num==2) { y=id; break; }
}
if (num!=2) { flag=0; break; }
vis[x]=vis[y]=1;
ans.push_back({b[pb][2],x,y});
++cnt;
}
if (!flag) { puts("No"); continue; }
puts("Yes");
for (auto [x,y,z]:ans) printf("%d %d %d\n",x,y,z);
}
return 0;
}
E. Corrupted Scoreboard Log
只能说太久不写代码是这样的，曾经手拿把掐的模拟题现在写的磕磕绊绊，最后还没判 00 的 Corner Case
这题本身没啥思维难度，把整个串分解后，考虑枚举第一题和前面的分界符
剩下每题直接分类讨论一下各种情况即可，注意到每个题的提交次数只有 \(1/2/3\) 位，三位的情况又很特殊可以直接判，因此这部分复杂度是 \(O(2^m)\) 的
#include<cstdio>
#include<iostream>
#include<string>
#include<vector>
#define RI register int
#define CI const int&
using namespace std;
int n,m; string ans; vector <string> vec; vector <int> tp; bool flag;
inline void DFS(CI tot_pro,CI tot_pnt,string ans,CI pro,CI pnt,CI now)
{
// if (tot_pnt==1322&&now<vec.size()) cout<<vec[now]<<endl;
// if (tot_pnt==1322) printf("now = %d,tot_pro = %d, tot_pnt = %d , pro = %d, pnt = %d\n",now,tot_pro,tot_pnt,pro,pnt);
if (now==vec.size())
{
if (pro==tot_pro&&pnt==tot_pnt)
{
cout<<ans<<endl; flag=1;
}
return;
}
string tmp_ans=ans;
if (tp[now]==1) // try case
{
if (vec[now]=="1")
{
ans+=" 1 try";
DFS(tot_pro,tot_pnt,ans,pro,pnt,now+1);
ans=tmp_ans;
return;
}
int cur_pnt=0;
for (RI j=0;j<vec[now].size()-1;++j)
cur_pnt=cur_pnt*10+vec[now][j]-'0';
if (vec[now][0]=='0')
{
if (cur_pnt==0&&vec[now].size()==2)
{
ans+=' ';
for (RI j=0;j<vec[now].size()-1;++j) ans+=vec[now][j];
ans+=' '; ans+=vec[now].back(); ans+=" try";
DFS(tot_pro,tot_pnt,ans,pro+1,pnt+cur_pnt,now+1);
ans=tmp_ans;
}
return;
}
if (cur_pnt>299) return;
ans+=' ';
for (RI j=0;j<vec[now].size()-1;++j) ans+=vec[now][j];
ans+=' '; ans+=vec[now].back(); ans+=" try";
DFS(tot_pro,tot_pnt,ans,pro+1,pnt+cur_pnt,now+1);
ans=tmp_ans;
return;
}
if (vec[now][0]!='0') //do not solve this problem
{
int try_times=0;
for (RI j=0;j<vec[now].size();++j)
try_times=try_times*10+vec[now][j]-'0';
if (try_times<=100)
{
ans+=' ';
for (RI j=0;j<vec[now].size();++j) ans+=vec[now][j];
ans+=" tries";
DFS(tot_pro,tot_pnt,ans,pro,pnt,now+1);
ans=tmp_ans;
}
}
if (flag) return;
if (vec[now].size()>=2&&vec[now].back()!='0'&&vec[now].back()!='1') //1 digit
{
int cur_pnt=0;
for (RI j=0;j<vec[now].size()-1;++j)
cur_pnt=cur_pnt*10+vec[now][j]-'0';
if (vec[now][0]=='0')
{
if (cur_pnt==0&&vec[now].size()==2)
{
ans+=' ';
for (RI j=0;j<vec[now].size()-1;++j) ans+=vec[now][j];
ans+=' '; ans+=vec[now].back(); ans+=" tries";
DFS(tot_pro,tot_pnt,ans,pro+1,pnt+cur_pnt+(vec[now].back()-'0'-1)*20,now+1);
ans=tmp_ans;
}
} else
if (cur_pnt<=299)
{
ans+=' ';
for (RI j=0;j<vec[now].size()-1;++j) ans+=vec[now][j];
ans+=' '; ans+=vec[now].back(); ans+=" tries";
DFS(tot_pro,tot_pnt,ans,pro+1,pnt+cur_pnt+(vec[now].back()-'0'-1)*20,now+1);
ans=tmp_ans;
}
}
if (flag) return;
if (vec[now].size()>=3&&vec[now][vec[now].size()-2]!='0') //2 digits
{
int cur_pnt=0;
for (RI j=0;j<vec[now].size()-2;++j)
cur_pnt=cur_pnt*10+vec[now][j]-'0';
int try_times=0;
for (RI j=vec[now].size()-2;j<vec[now].size();++j)
try_times=try_times*10+vec[now][j]-'0';
if (vec[now][0]=='0')
{
if (cur_pnt==0&&vec[now].size()==3)
{
ans+=' ';
for (RI j=0;j<vec[now].size()-2;++j) ans+=vec[now][j];
ans+=' ';
for (RI j=vec[now].size()-2;j<vec[now].size();++j) ans+=vec[now][j];
ans+=" tries";
DFS(tot_pro,tot_pnt,ans,pro+1,pnt+cur_pnt+(try_times-1)*20,now+1);
ans=tmp_ans;
}
} else
if (cur_pnt<=299)
{
ans+=' ';
for (RI j=0;j<vec[now].size()-2;++j) ans+=vec[now][j];
ans+=' ';
for (RI j=vec[now].size()-2;j<vec[now].size();++j) ans+=vec[now][j];
ans+=" tries";
DFS(tot_pro,tot_pnt,ans,pro+1,pnt+cur_pnt+(try_times-1)*20,now+1);
ans=tmp_ans;
}
}
if (flag) return;
if (vec[now].size()>=4&&vec[now][vec[now].size()-3]=='1'&&vec[now][vec[now].size()-2]=='0'&&vec[now].back()=='0') // 3 digits (only 100)
{
int cur_pnt=0;
for (RI j=0;j<vec[now].size()-3;++j)
cur_pnt=cur_pnt*10+vec[now][j]-'0';
if (vec[now][0]=='0')
{
if (cur_pnt==0&&vec[now].size()==4)
{
ans+=' ';
for (RI j=0;j<vec[now].size()-3;++j) ans+=vec[now][j];
ans+=' ';
ans+="100 tries";
DFS(tot_pro,tot_pnt,ans,pro+1,pnt+cur_pnt+(100-1)*20,now+1);
ans=tmp_ans;
}
} else
if (cur_pnt<=299)
{
ans+=' ';
for (RI j=0;j<vec[now].size()-3;++j) ans+=vec[now][j];
ans+=' ';
ans+="100 tries";
DFS(tot_pro,tot_pnt,ans,pro+1,pnt+cur_pnt+(100-1)*20,now+1);
ans=tmp_ans;
}
}
if (flag) return;
}
inline void solve(CI tot_pro)
{
// printf("tot_pro = %d\n",tot_pro);
for (RI i=1;i<=6;++i)
{
string tmp_ans=ans;
if (i>=vec[0].size()) break;
string tmp=vec[0];
// cout<<"tmp = "<<tmp<<endl;
int tot_pnt=0;
for (RI j=0;j<tmp.size()-i;++j)
tot_pnt=tot_pnt*10+tmp[j]-'0';
if (tmp[0]=='0')
{
if (tmp.size()-i!=1) continue;
}
vec[0]="";
for (RI j=tmp.size()-i;j<tmp.size();++j)
vec[0]+=tmp[j];
for (RI j=0;j<tmp.size()-i;++j) ans+=tmp[j];
// cout<<"i = "<<i<<" "<<vec[0]<<endl;
DFS(tot_pro,tot_pnt,ans,0,0,0);
if (flag) return;
vec[0]=tmp;
ans=tmp_ans;
}
}
int main()
{
ios::sync_with_stdio(0); cin.tie(0);
cin>>n>>m;
for (RI id=1;id<=n;++id)
{
vec.clear(); tp.clear(); flag=0;
string s; cin>>s; string tmp="";
if (s=="00") { cout<<"0 0"<<endl; continue; }
for (RI i=0;i<s.size();++i)
{
if (s[i]=='t')
{
vec.push_back(tmp); tmp="";
if (s[i+2]=='y') tp.push_back(1),i+=2;
else tp.push_back(0),i+=4;
continue;
}
tmp+=s[i];
}
int d=vec[0][0]-'0';
vec[0].erase(vec[0].begin());
ans=""; ans+=char(d+'0'); ans+=' '; solve(d);
if (flag) continue;
int dd=vec[0][0]-'0';
vec[0].erase(vec[0].begin());
ans=""; ans+=char(d+'0'); ans+=char(dd+'0'); ans+=' '; solve(d*10+dd);
}
return 0;
}
F. Boolean Function Reconstruction
被徐神单切力
考虑一种很 trivial 的构造方式，对于所有 \(f(i)=1\) 的状态 \(i\)，直接把它对应的与表达式或到最后的答案中即可
更进一步的观察，我们可以发现由于序列中没有取反，因此如果状态 \(f(x)=1\)，那么 \(x\) 的所有超集 \(y\) 都满足 \(f(y)=1\)，因此可以贪心地构造最小的那些表示状态
但如果直接这么做还是会超过符号限制，因此我们不妨合并一些子项，即按顺序提取出每一个字符，再将剩下的子项按照含有/不含有这个字符分类，递归构造即可
#include <bits/stdc++.h>
namespace check {
using bitset = std::bitset<1 << 15>;
bitset val[15];
bitset eval(const char *&s) {
if(isalpha(*s)) return val[*s++ - 'a'];
assert(*s == '(');
auto val1 = eval(++s);
const char *p1 = s;
auto val2 = eval(++s);
assert(*s++ == ')');
if(*p1 == '&') return val1 & val2;
else return val1 | val2;
}
std::string check(int n, std::string s) {
for(int i = 0; i < n; ++i) for(int j = 0; j < (1 << n); ++j)
val[i][j] = (j >> i & 1);
const char *p = s.c_str();
auto result = eval(p);
std::stringstream res;
for(int i = 0; i < (1 << n); ++i) res << result[i];
return res.str();
}
}
void generate(int i, const std::vector<int> &s, std::stringstream &ss) {
std::vector<int> a, b;
int aa = 0;
for(auto s: s) {
if(s >> i & 1) {
if(s == (1 << i)) aa = 1;
else a.push_back(s ^ (1 << i));
} else b.push_back(s);
}
if((a.size() || aa) && b.size()) ss << '(';
if(a.size()) ss << '(';
if(aa || a.size()) ss << char(i + 'a');
if(a.size()) ss << '&', generate(i + 1, a, ss), ss << ')';
if((a.size() || aa) && b.size()) ss << '|';
if(b.size()) generate(i + 1, b, ss);
if((a.size() || aa) && b.size()) ss << ')';
return ;
}
void work() {
int n; std::cin >> n;
std::string input;
std::vector<bool> c(1 << n); {
std::cin >> input;
for(int i = 0; i < (1 << n); ++i) c[i] = (input[i] == '1');
}
int op_count = 0;
int count0 = 0;
for(int i = 0; i < (1 << n); ++i) count0 += (c[i] == 0);
if(count0 == 0 ) return std::cout << "Yes\nT\n", void(0);
if(count0 == (1 << n)) return std::cout << "Yes\nF\n", void(0);
if(c[0] == 1) return std::cout << "No\n", void(0);
std::vector<bool> d(1 << n, 0);
std::vector<int> ans;
for(int i = 1; i < (1 << n); ++i) {
if(d[i] && !c[i]) return std::cout << "No\n", void(0);
if(d[i] == c[i]) continue;
ans.push_back(i);
for(int S = ((1 << n) - 1) ^ i; ; S = (((1 << n) - 1) ^ i) & (S - 1)) {
d[S | i] = 1;
if(S == 0) break;
}
}
std::cout << "Yes\n";
std::stringstream anss;
generate(0, ans, anss);
std::cout << anss.str() << char(10);
return ;
}
int main() {
// freopen("data.in", "r", stdin);
// freopen("data.out", "w", stdout);
std::ios::sync_with_stdio(false);
int T; std::cin >> T; while(T--) work();
fclose(stdin);
return 0;
}
G. Collatz Conjecture
首先注意到 \([1,B]\) 中的所有数都一定在某个环上
这是因为对于某个 \(x\in [1,B]\)，它在加上了若干次 \(B\) 并成为 \(y\times A\) 后，除去 \(A\) 变为的 \(y\) 一定也 \(\in[1,B]\)
不妨假设初始的 \(n\) 在环上，则它一定由某个 \(r\in[1,B]\) 不断加 \(B\) 得来，很容易发现这个 \(r\) 和 \(n\) 模 \(B\) 意义下同余
定下 \(r\) 后判断 \(n\) 是否合法只要找出环上最大的值即可，设进行了 \(x\) 次加 \(B\) 操作，则有 \(r+x\times B=y\times A\)，用 exgcd 解出最小的 \(x\) 即可
#include<bits/stdc++.h>
using namespace std;
#define int long long
int ex_gcd(int a, int b, int &x, int &y) {
if (0 == b) {x = 1; y = 0; return a;}
int ret = ex_gcd(b, a%b, y, x);
y -= a / b * x;
return ret;
}
int calc(int A, int B, int R) { //xA + yB = R 的最小正整数 x
int x0, y0;
int g = ex_gcd(A, B, x0, y0);
int x00 = x0 * (R / g), y00 = y0 * (R / g);
int x1 = B / g, y1 = A / g;
return ((x00 % x1) + x1) % x1;
}
bool solve() {
int A, B, n; cin >> A >> B >> n;
int r = n % B;
if (0 == r) r = B;
int x = calc(B, A, -r);
// printf("x = %d\n", x);
int n0 = r + x * B;
return n <= n0;
}
signed main() {
ios::sync_with_stdio(0); cin.tie(0);
int T; cin >> T; while (T--) cout << (solve() ? "Yes\n" : "No\n");
return 0;
}
H. Staircase Museum
我写 E 的时候队友把这题讨论出来了，但后面因为时间原因也没写完
赛后看了下队友的写法好像和官方题解不太一样，感觉很神奇
#include <bits/stdc++.h>
using namespace std;
const int INF = (int)1e9+5;
#define ft first
#define sd second
const int N = 2e6+50;
int l[N], r[N], yyy[4][N], dp[N], deg[N];
vector<int> ul, dr;
vector<int> G[N];
map<pair<int, int>, int> id; int cntid=0;
struct Node {
int x, y, typ;
};
vector<Node> que; int ed=-1, fr=0;
// int allocate(pair<int, int> pr) {
// if (!id.count(pr)) id[pr] = ++cntid;
// return id[pr];
// }
int getdist() {
vector<int> que2; int ed2=-1, fr2=0;
for (int i=0; i<=cntid; ++i) {
if (deg[i]==0) que2.push_back(i), ++ed2;
}
while (fr2 <= ed2) {
int x = que2[fr2++];
// printf("x=%d fr2=%d ed2=%d\n", x, fr2, ed2);
for (int v : G[x]) if (deg[v]>0){
// printf("v=%d\n", v);
dp[v] = max(dp[v], dp[x]+1);
--deg[v];
if (0 == deg[v]) que2.push_back(v), ++ed2;
}
}
return dp[0];
}
int solve() {
int n; cin >> n;
for (int i=1; i<=n; ++i) cin >> l[i] >> r[i];
for (int i=0; i<=n; ++i) {
if (i+1 <= n) yyy[0][i] = r[i+1]; else yyy[0][i] = INF;
if (i > 0) yyy[1][i] = r[i]; else yyy[1][i] = r[1];
if (i+1 <= n) yyy[2][i] = l[i+1]-1; else yyy[0][i] = INF;
if (i > 0) yyy[3][i] = l[i]-1; else yyy[3][i] = l[1];
}
ul.push_back(-1);
dr.push_back(-1);
ul.push_back(0);
for (int i=1; i<n; ++i) {
if (yyy[0][i] != yyy[0][i-1]) ul.push_back(i);
}
for (int i=1; i<n; ++i) {
if (yyy[3][i] != yyy[3][i+1]) dr.push_back(i);
}
dr.push_back(n);
auto pr1 = make_pair(ul.back(), yyy[0][ul.back()]);
// printf("ul:(%d %d)\n", ul.back(), yyy[0][ul.back()]);
que.push_back(Node{pr1.ft, pr1.sd, 0}); ++ed;
id[pr1] = ++cntid;
dp[cntid] = 0;
auto pr2 = make_pair(dr.back(), yyy[3][dr.back()]);
// printf("dr:(%d %d)\n", dr.back(), yyy[3][dr.back()]);
que.push_back(Node{pr2.ft, pr2.sd, 1}); ++ed;
id[pr2] = ++cntid;
dp[cntid] = 0;
while (fr <= ed) {
auto [x, y, typ] = que[fr++];
int cur = id[make_pair(x, y)];
// printf("x=%d y=%d typ=%d\n", x, y, typ);
if (0 == typ) {
if (x > 0) {
int x1 = x;
int y1 = yyy[3][x];
int tmpid;
if (!id.count({x1, y1})) {
que.push_back(Node{x1, y1, 1}); ++ed;
id[{x1, y1}] = ++cntid;
tmpid = cntid;
} else {
tmpid = id[{x1, y1}];
}
G[cur].push_back(tmpid);
++deg[tmpid];
} else {
G[cur].push_back(0);
++deg[0];
}
int x0 = *prev(lower_bound(ul.begin(), ul.end(), x));
// printf("x0=%d\n", x0);
if (x0 > -1) {
int y0 = yyy[0][x0];
int tmpid;
if (!id.count({x0, y0})) {
que.push_back(Node{x0, y0, 0}); ++ed;
id[{x0, y0}] = ++cntid;
tmpid = cntid;
} else {
tmpid = id[{x0, y0}];
}
G[cur].push_back(tmpid);
++deg[tmpid];
}
} else {
if (y > l[1]-1) {
int x0 = lower_bound(yyy[0], yyy[0]+n+1, y) - yyy[0];
int y0 = y;
int tmpid;
if (!id.count({x0, y0})) {
que.push_back(Node{x0, y0, 0}); ++ed;
id[{x0, y0}] = ++cntid;
tmpid = cntid;
} else {
tmpid = id[{x0, y0}];
}
G[cur].push_back(tmpid);
++deg[tmpid];
} else {
G[cur].push_back(0);
++deg[0];
}
int x1 = *prev(lower_bound(dr.begin(), dr.end(), x));
if (x1 > -1) {
int y1 = yyy[3][x1];
int tmpid;
if (!id.count({x1, y1})) {
que.push_back(Node{x1, y1, 1}); ++ed;
id[{x1, y1}] = ++cntid;
tmpid = cntid;
} else {
tmpid = id[{x1, y1}];
}
G[cur].push_back(tmpid);
++deg[tmpid];
}
}
}
// for (int i=1; i<=cntid; ++i) {
// for (int v : G[i]) {
// dp[v] = max(dp[v], dp[i]+1);
// }
// }
int ans = getdist();
// printf("y1:"); for (int i=0; i<=n; ++i) printf("%d ", yyy[1][i]); puts("");
// for (auto [pri, idd] : id) {
// printf("(%d %d) %d\n", pri.ft, pri.sd, idd);
// }
// for (int i=1; i<=cntid; ++i) {
// printf("%d:", i);
// for (int v : G[i]) {
// printf("%d ", v);
// }puts("");
// }
// printf("dp:"); for (int i=0; i<=cntid; ++i) printf("%d ", dp[i]); puts("");
for (int i=0; i<=cntid; ++i) {
dp[i] = -1;
deg[i] = 0;
G[i].clear();
}
ul.clear(); dr.clear();
id.clear(); cntid = 0;
ed = -1, fr = 0; que.clear();
return ans;
}
signed main() {
ios::sync_with_stdio(0); cin.tie(0);
int T; cin >> T; while (T--) cout << solve() << '\n';
return 0;
}
I. Color-Balanced Tree
应该也是个签到构造，我题目都没看
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+5;
int n, col[N], cnt[2];
int ed=-1, fr=0, que[N];
vector<int> G[N];
void solve() {
cin >> n;
ed = -1, fr = 0;
cnt[0] = cnt[1] = 0;
for (int i=1; i<=2*n; ++i) {
col[i] = -1;
G[i].clear();
}
for (int i=1; i<2*n; ++i) {
int u, v; cin >> u >> v;
G[u].push_back(v);
G[v].push_back(u);
}
col[1] = 1;
++cnt[1];
que[++ed] = 1;
while (fr<=ed) {
int x = que[fr++];
bool isleaf = true;
for (int v : G[x]) if (-1 == col[v]) {
col[v] = col[x]^1;
++cnt[col[v]];
que[++ed] = v;
isleaf = false;
}
if (isleaf) {
--cnt[col[x]];
col[x] = -1;
}
}
// printf("n=%d\n", n);
for (int i=1; i<=2*n; ++i) {
if (-1==col[i]) {
if (cnt[0] <= cnt[1]) {
++cnt[0];
col[i] = 0;
} else {
++cnt[1];
col[i] = 1;
}
}
cout << col[i] << (i==2*n ? '\n' : ' ');
}
}
Postscript
五一训练保三争四吧，不然现在的水平和状态包去被薄纱的
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