A football league has recently begun in Beautiful land. There are $n$ teams participating in the league. Let’s enumerate them with integers from $1$ to $n$.
There will be played exactly $\frac{n(n-1)}{2}$ matches: each team will play against all other teams exactly once. In each match, there is always a winner and loser and there is no draw.
After all matches are played, the organizers will count the number of beautiful triples. Let’s call a triple of three teams $(A, B, C)$ beautiful if a team $A$ win against a team $B$, a team $B$ win against a team $C$ and a team $C$ win against a team $A$. We look only to a triples of different teams and the order of teams in the triple is important.
The beauty of the league is the number of beautiful triples.
At the moment, $m$ matches were played and their results are known.
What is the maximum beauty of the league that can be, after playing all remaining matches? Also find a possible results for all remaining $\frac{n(n-1)}{2} – m$ matches, so that the league has this maximum beauty.Input
The first line contains two integers $n, m$ ($3 \leq n \leq 50, 0 \leq m \leq \frac{n(n-1)}{2}$) — the number of teams in the football league and the number of matches that were played.
Each of $m$ following lines contains two integers $u$ and $v$ ($1 \leq u, v \leq n$, $u \neq v$) denoting that the $u$-th team won against the $v$-th team. It is guaranteed that each unordered pair of teams appears at most once.Output
Print $n$ lines, each line having a string of exactly $n$ characters. Each character must be either $0$ or $1$.
Let $a_{ij}$ be the $j$-th number in the $i$-th line. For all $1 \leq i \leq n$ it should be true, that $a_{ii} = 0$. For all pairs of teams $i \neq j$ the number $a_{ij}$ indicates the result of the match between the $i$-th team and the $j$-th team:
- If $a_{ij}$ is $1$, the $i$-th team wins against the $j$-th team;
- Otherwise the $j$-th team wins against the $i$-th team;
- Also, it should be true, that $a_{ij} + a_{ji} = 1$.
Also note that the results of the $m$ matches that were already played cannot be changed in your league.
The beauty of the league in the output should be maximum possible. If there are multiple possible answers with maximum beauty, you can print any of them.Examplesinput
3 1 1 2
output
010 001 100
input
4 2 1 2 1 3
output
0110 0001 0100 1010
Note
The beauty of league in the first test case is equal to $3$ because there exists three beautiful triples: $(1, 2, 3)$, $(2, 3, 1)$, $(3, 1, 2)$.
The beauty of league in the second test is equal to $6$ because there exists six beautiful triples: $(1, 2, 4)$, $(2, 4, 1)$, $(4, 1, 2)$, $(2, 4, 3)$, $(4, 3, 2)$, $(3, 2, 4)$.
Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
string to_string(pair<A, B> p);
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);
string to_string(const string& s) {
return '"' + s + '"';
}
string to_string(const char* s) {
return to_string((string) s);
}
string to_string(bool b) {
return (b ? "true" : "false");
}
string to_string(vector<bool> v) {
bool first = true;
string res = "{";
for (int i = 0; i < static_cast<int>(v.size()); i++) {
if (!first) {
res += ", ";
}
first = false;
res += to_string(v[i]);
}
res += "}";
return res;
}
template <size_t N>
string to_string(bitset<N> v) {
string res = "";
for (size_t i = 0; i < N; i++) {
res += static_cast<char>('0' + v[i]);
}
return res;
}
template <typename A>
string to_string(A v) {
bool first = true;
string res = "{";
for (const auto &x : v) {
if (!first) {
res += ", ";
}
first = false;
res += to_string(x);
}
res += "}";
return res;
}
template <typename A, typename B>
string to_string(pair<A, B> p) {
return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ")";
}
template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
cerr << " " << to_string(H);
debug_out(T...);
}
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif
template <typename T, typename C>
class mcmf {
public:
static constexpr T eps = (T) 1e-9;
struct edge {
int from;
int to;
T c;
T f;
C cost;
};
vector<vector<int>> g;
vector<edge> edges;
vector<C> d;
vector<int> q;
vector<bool> in_queue;
vector<int> pe;
int n;
int st, fin;
T flow;
C cost;
mcmf(int _n, int _st, int _fin) : n(_n), st(_st), fin(_fin) {
assert(0 <= st && st < n && 0 <= fin && fin < n && st != fin);
g.resize(n);
d.resize(n);
in_queue.resize(n);
pe.resize(n);
flow = 0;
cost = 0;
}
void clear_flow() {
for (const edge &e : edges) {
e.f = 0;
}
flow = 0;
}
void add(int from, int to, T forward_cap, T backward_cap, C cost) {
assert(0 <= from && from < n && 0 <= to && to < n);
g[from].push_back((int) edges.size());
edges.push_back({from, to, forward_cap, 0, cost});
g[to].push_back((int) edges.size());
edges.push_back({to, from, backward_cap, 0, -cost});
}
bool expath() {
fill(d.begin(), d.end(), numeric_limits<C>::max());
q.clear();
q.push_back(st);
d[st] = 0;
in_queue[st] = true;
int beg = 0;
bool found = false;
while (beg < (int) q.size()) {
int i = q[beg++];
if (i == fin) {
found = true;
}
in_queue[i] = false;
for (int id : g[i]) {
const edge &e = edges[id];
if (e.c - e.f > eps && d[i] + e.cost < d[e.to]) {
d[e.to] = d[i] + e.cost;
pe[e.to] = id;
if (!in_queue[e.to]) {
q.push_back(e.to);
in_queue[e.to] = true;
}
}
}
}
if (found) {
T push = numeric_limits<T>::max();
int v = fin;
while (v != st) {
const edge &e = edges[pe[v]];
push = min(push, e.c - e.f);
v = e.from;
}
v = fin;
while (v != st) {
edge &e = edges[pe[v]];
e.f += push;
edge &back = edges[pe[v] ^ 1];
back.f -= push;
v = e.from;
}
flow += push;
cost += push * d[fin];
}
return found;
}
pair<T, C> max_flow_min_cost() {
while (expath()) {}
return {flow, cost};
}
};
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
int n, m;
cin >> n >> m;
vector<string> g(n, string(n, '0'));
vector<int> deg(n);
for (int i = 0; i < m; i++) {
int x, y;
cin >> x >> y;
--x; --y;
g[x][y] = '1';
++deg[x];
}
int rm = n * (n - 1) / 2 - m;
mcmf<int, int> f(n + rm + 2, n + rm, n + rm + 1);
for (int i = 0; i < n; i++) {
for (int j = deg[i] + 1; j <= n - 1; j++) {
f.add(f.st, i, 1, 0, j * (j - 1) / 2 - (j - 1) * (j - 2) / 2);
}
}
vector<pair<int, int>> who;
int id = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (g[i][j] == '0' && g[j][i] == '0') {
f.add(i, n + id, 1, 0, 0);
f.add(j, n + id, 1, 0, 0);
f.add(n + id, f.fin, 1, 0, 0);
who.emplace_back(i, j);
++id;
}
}
}
f.max_flow_min_cost();
for (int i = 0; i < n; i++) {
for (int eid : f.g[i]) {
if (f.edges[eid].f == 1) {
int id = f.edges[eid].to - n;
int j = who[id].first ^ who[id].second ^ i;
g[i][j] = '1';
}
}
}
for (int i = 0; i < n; i++) {
cout << g[i] << '\n';
}
return 0;
}
