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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#854612#9733. Heavy-light Decompositionucup-team3099#WA 3ms3772kbC++204.3kb2025-01-12 03:29:392025-01-12 03:29:39

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你现在查看的是最新测评结果

  • [2025-01-12 03:29:39]
  • 评测
  • 测评结果:WA
  • 用时:3ms
  • 内存:3772kb
  • [2025-01-12 03:29:39]
  • 提交

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

Tip: Click on the bar to expand more detailed information

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)