結果

問題 No.2263 Perms
ユーザー fuppy_kyopro
提出日時 2023-04-07 21:56:13
言語 C++17(gcc12)
(gcc 12.3.0 + boost 1.87.0)
結果
AC  
実行時間 108 ms / 2,000 ms
コード長 4,931 bytes
コンパイル時間 13,976 ms
コンパイル使用メモリ 300,928 KB
最終ジャッジ日時 2025-02-12 01:14:57
ジャッジサーバーID
(参考情報)
judge2 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 39
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

//*
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
//*/
// #include <atcoder/all>
#include <atcoder/maxflow>
#include <bits/stdc++.h>
using namespace std;
using namespace atcoder;
// #define _GLIBCXX_DEBUG
#define DEBUG(x) cerr << #x << ": " << x << endl;
#define DEBUG_VEC(v) \
cerr << #v << ":"; \
for (int iiiiiiii = 0; iiiiiiii < v.size(); iiiiiiii++) \
cerr << " " << v[iiiiiiii]; \
cerr << endl;
#define DEBUG_MAT(v) \
cerr << #v << endl; \
for (int iv = 0; iv < v.size(); iv++) { \
for (int jv = 0; jv < v[iv].size(); jv++) { \
cerr << v[iv][jv] << " "; \
} \
cerr << endl; \
}
typedef long long ll;
// #define int ll
#define vi vector<int>
#define vl vector<ll>
#define vii vector<vector<int>>
#define vll vector<vector<ll>>
#define vs vector<string>
#define pii pair<int, int>
#define pis pair<int, string>
#define psi pair<string, int>
#define pll pair<ll, ll>
template <class S, class T>
pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) {
return pair<S, T>(s.first + t.first, s.second + t.second);
}
template <class S, class T>
pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template <class S, class T>
ostream &operator<<(ostream &os, pair<S, T> p) {
os << "(" << p.first << ", " << p.second << ")";
return os;
}
#define X first
#define Y second
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep1(i, n) for (int i = 1; i <= (int)(n); i++)
#define rrep(i, n) for (int i = (int)(n)-1; i >= 0; i--)
#define rrep1(i, n) for (int i = (int)(n); i > 0; i--)
#define REP(i, a, b) for (int i = a; i < b; i++)
#define in(x, a, b) (a <= x && x < b)
#define all(c) c.begin(), c.end()
void YES(bool t = true) {
cout << (t ? "YES" : "NO") << endl;
}
void Yes(bool t = true) { cout << (t ? "Yes" : "No") << endl; }
void yes(bool t = true) { cout << (t ? "yes" : "no") << endl; }
void NO(bool t = true) { cout << (t ? "NO" : "YES") << endl; }
void No(bool t = true) { cout << (t ? "No" : "Yes") << endl; }
void no(bool t = true) { cout << (t ? "no" : "yes") << endl; }
template <class T>
bool chmax(T &a, const T &b) {
if (a < b) {
a = b;
return 1;
}
return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
if (a > b) {
a = b;
return 1;
}
return 0;
}
#define UNIQUE(v) v.erase(std::unique(v.begin(), v.end()), v.end());
const ll inf = 1000000001;
const ll INF = (ll)1e18 + 1;
const long double pi = 3.1415926535897932384626433832795028841971L;
int popcount(ll t) { return __builtin_popcountll(t); }
// int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
// int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
vi dx = {0, 0, -1, 1}, dy = {-1, 1, 0, 0};
vi dx2 = {1, 1, 0, -1, -1, -1, 0, 1}, dy2 = {0, 1, 1, 1, 0, -1, -1, -1};
struct Setup_io {
Setup_io() {
ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
cout << fixed << setprecision(25);
}
} setup_io;
// const ll MOD = 1000000007;
const ll MOD = 998244353;
// #define mp make_pair
//#define endl '\n'
vi solve(int n, vii a) {
int V = 1 + n + n + 1;
int S = V - 2, T = V - 1;
mf_graph<int> g(V);
vector<pii> edges;
rep(i, n) {
rep(x, n) {
if (a[i][x]) {
g.add_edge(x, n + i, 1);
edges.emplace_back(x, i);
}
}
}
rep(x, n) {
g.add_edge(S, x, 1);
}
rep(i, n) {
g.add_edge(n + i, T, 1);
}
int f = g.flow(S, T);
if (f != n) {
cout << -1 << endl;
exit(0);
}
vi p(n);
rep(i, edges.size()) {
auto e = g.get_edge(i);
if (e.flow == 1) {
auto [x, idx] = edges[i];
p[idx] = x;
}
}
return p;
}
signed main() {
int n, m;
cin >> n >> m;
vii a(n, vi(n));
rep(i, n) {
rep(j, n) {
cin >> a[i][j];
}
}
rep(i, n) {
int s = 0;
rep(j, n) {
s += a[i][j];
}
if (s != m) {
cout << -1 << endl;
return 0;
}
s = 0;
rep(j, n) {
s += a[j][i];
}
if (s != m) {
cout << -1 << endl;
return 0;
}
}
vii ans(m);
rep(i, m) {
vi ret = solve(n, a);
ans[i] = ret;
rep(j, ret.size()) {
a[j][ret[j]]--;
}
}
rep(i, m) {
for (int x : ans[i]) {
cout << x + 1 << " ";
}
cout << endl;
}
}
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