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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#854612 | #9733. Heavy-light Decomposition | ucup-team3099# | WA | 3ms | 3772kb | C++20 | 4.3kb | 2025-01-12 03:29:39 | 2025-01-12 03:29:39 |
Judging History
answer
#ifdef LOCAL
#define _GLIBCXX_DEBUG 1
#define dbg(...) cerr << "LINE(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__)
#else
#define dbg(...) 0
#endif
#if 0
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
template<class T>
using ordered_set = __gnu_pbds::tree<T, __gnu_pbds::null_type, std::less<T>, __gnu_pbds::rb_tree_tag,
__gnu_pbds::tree_order_statistics_node_update>;
#endif
#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <random>
#include <chrono>
#include <cassert>
using namespace std;
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define sz(x) (int)(x).size()
#define all(x) begin(x), end(x)
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define F0R(i,a) FOR(i,0,a)
#define REP(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define each(a,x) for (auto& a: x)
#define tcT template<class T
#define tcTU tcT, class U
#define tcTUU tcT, class ...U
template<class T> using V = vector<T>;
template<class T, size_t SZ> using AR = array<T,SZ>;
typedef string str;
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
template<typename T, typename U> T &ctmax(T &x, const U &y){ return x = max<T>(x, y); }
template<typename T, typename U> T &ctmin(T &x, const U &y){ return x = min<T>(x, y); }
mt19937 rng((unsigned)chrono::steady_clock::now().time_since_epoch().count());
#define ts to_string
str ts(char c) { return str(1,c); }
str ts(bool b) { return b ? "true" : "false"; }
str ts(const char* s) { return (str)s; }
str ts(str s) { return s; }
str ts(vector<bool> v) { str res = "{"; F0R(i,sz(v)) res += char('0'+v[i]); res += "}"; return res; }
template<size_t SZ> str ts(bitset<SZ> b) { str res = ""; F0R(i,SZ) res += char('0'+b[i]); return res; }
template<class A, class B> str ts(pair<A,B> p);
template<class T> str ts(T v) { bool fst = 1; str res = "{"; for (const auto& x: v) {if (!fst) res += ", "; fst = 0; res += ts(x);} res += "}"; return res;}
template<class A, class B> str ts(pair<A,B> p) {return "("+ts(p.first)+", "+ts(p.second)+")"; }
template<class A> void pr(A x) { cout << ts(x); }
template<class H, class... T> void pr(const H& h, const T&... t) { pr(h); pr(t...); }
void ps() { pr("\n"); }
template<class H, class... T> void ps(const H& h, const T&... t) { pr(h); if (sizeof...(t)) pr(" "); ps(t...); }
void DBG() { cerr << "]" << endl; }
template<class H, class... T> void DBG(H h, T... t) {cerr << ts(h); if (sizeof...(t)) cerr << ", "; DBG(t...); }
tcTU> void re(pair<T,U>& p);
tcT> void re(V<T>& v);
tcT, size_t SZ> void re(AR<T,SZ>& a);
tcT> void re(T& x) { cin >> x; }
void re(double& d) { str t; re(t); d = stod(t); }
void re(long double& d) { str t; re(t); d = stold(t); }
tcTUU> void re(T& t, U&... u) { re(t); re(u...); }
tcTU> void re(pair<T,U>& p) { re(p.first,p.second); }
tcT> void re(V<T>& x) { each(a,x) re(a); }
tcT, size_t SZ> void re(AR<T,SZ>& x) { each(a,x) re(a); }
tcT> void rv(int n, V<T>& x) { x.rsz(n); re(x); }
constexpr bool multitest() {return 1;}
void solve();
int main() {
ios_base::sync_with_stdio(false); cin.tie(NULL);
int t = 1;
if (multitest()) cin >> t;
for (; t; t--) solve();
}
void solve() {
int n, k; re(n, k);
vector<pii> chains(k); re(chains);
vi p(n+1);
for (auto [l, r] : chains) {
for (int i = l+1; i <= r; i++) p[i] = i-1;
}
sort(all(chains), [&](pii a, pii b) {return a.second-a.first > b.second-b.first;});
if (k > 1 && chains[0].second - chains[0].first == chains.back().second - chains.back().first) {
ps("IMPOSSIBLE");
return;
}
for (int i = 1; i < k; i++) {
p[chains[i].first] = max(chains[0].first, chains[0].second - (chains[i].second-chains[i].first+1));
}
for (int i = 1; i <= n; i++) pr(p[i], " ");
ps();
}
Details
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Test #1:
score: 100
Accepted
time: 0ms
memory: 3712kb
input:
3 12 5 1 5 9 11 7 8 6 6 12 12 4 3 1 1 4 4 2 3 2 2 1 1 2 2
output:
0 1 2 3 4 4 3 7 2 9 10 4 2 0 2 2 IMPOSSIBLE
result:
ok Correct. (3 test cases)
Test #2:
score: -100
Wrong Answer
time: 3ms
memory: 3772kb
input:
10 1 1 1 1 100000 1 1 100000 12 4 1 3 4 6 7 9 10 12 6 3 4 6 2 3 1 1 8999 3 1 3000 3001 6000 6001 8999 14 4 1 3 4 6 7 10 11 14 17 5 1 3 4 6 7 10 11 14 15 17 19999 2 1 9999 10000 19999 1 1 1 1 5 3 1 1 2 3 4 5
output:
0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101...
result:
wrong answer 2 should be a heavy child, but the maximum one with size 3000 and 2999 found. (test case 5)