AtCoder Beginner Contest 317 G Rearranging
考虑转化成匹配问题。
对每行建点 \(p_i\)，每个数建点 \(q_i\)，\(\forall i \in [1, n], j \in [1, m]\)，连边 \((p_i, q_{a_{i, j}})\)，问题转化成求这个二分图能否恰好被分解成 \(m\) 组完美匹配以及方案。
直接给出做法：每次任意找出一组完美匹配，并把边从图中删除，做 \(m\) 次。若有一次找不出完美匹配就无解。
为什么是对的呢？
考虑 Hall 定理。二分图有完美匹配的充要条件是，对于左部点集的任意一个子集 \(S\)，它的邻居集合 \(N(S)\)（可重集）均满足 \(|S| \le |N(S)|\)。
Hall 定理是可以推广的，二分图有 \(m\) 组完美匹配的充要条件是，对于左部点集的任意一个子集 \(S\)，它的邻居集合 \(N(S)\) 均满足 \(m|S| \le |N(S)|\)。
如果原图满足这个条件，那么随便删除一组完美匹配后也依然满足。如果原图不满足这个条件，那么它就找不出来 \(m\) 组完美匹配。
使用 Dinic 求二分图匹配，复杂度 \(O(n^{1.5}m^2)\)。
code
// Problem: G - Rearranging
// Contest: AtCoder - GAMEFREAK Programming Contest 2023 (AtCoder Beginner Contest 317)
// URL: https://atcoder.jp/contests/abc317/tasks/abc317_g
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;
const int maxn = 110;
const int maxm = 1000000;
const int inf = 0x3f3f3f3f;
int n, m, tot, a[maxn][maxn], b[maxn][maxn], c[maxn][maxn];
int id1[maxn], id2[maxn], ntot, head[maxm], len = 1, S, T;
vector<int> vc[maxn][maxn];
struct E {
int u, v, d;
E(int a = 0, int b = 0, int c = 0) : u(a), v(b), d(c) {}
} G[maxm];
struct edge {
int to, next, cap, flow, id;
} edges[maxm];
inline void add_edge(int u, int v, int c, int f, int id) {
edges[++len].to = v;
edges[len].next = head[u];
edges[len].cap = c;
edges[len].flow = f;
edges[len].id = id;
head[u] = len;
}
struct Dinic {
int d[maxm], cur[maxm];
bool vis[maxm];
inline void add(int u, int v, int c, int id) {
add_edge(u, v, c, 0, id);
add_edge(v, u, 0, 0, id);
}
inline bool bfs() {
for (int i = 1; i <= ntot; ++i) {
vis[i] = 0;
d[i] = -1;
}
queue<int> q;
vis[S] = 1;
d[S] = 0;
q.push(S);
while (q.size()) {
int u = q.front();
q.pop();
for (int i = head[u]; i; i = edges[i].next) {
edge &e = edges[i];
if (!vis[e.to] && e.cap > e.flow) {
vis[e.to] = 1;
d[e.to] = d[u] + 1;
q.push(e.to);
}
}
}
return vis[T];
}
int dfs(int u, int a) {
if (u == T || !a) {
return a;
}
int flow = 0, f;
for (int &i = cur[u]; i; i = edges[i].next) {
edge &e = edges[i];
if (d[e.to] == d[u] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {
e.flow += f;
edges[i ^ 1].flow -= f;
flow += f;
a -= f;
if (!a) {
break;
}
}
}
return flow;
}
inline int solve() {
int flow = 0;
while (bfs()) {
for (int i = 1; i <= ntot; ++i) {
cur[i] = head[i];
}
flow += dfs(S, inf);
}
return flow;
}
} solver;
void solve() {
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) {
id1[i] = ++ntot;
for (int j = 1; j <= m; ++j) {
scanf("%d", &a[i][j]);
++c[i][a[i][j]];
}
}
for (int i = 1; i <= n; ++i) {
id2[i] = ++ntot;
}
S = ++ntot;
T = ++ntot;
for (int i = 1; i <= n; ++i) {
G[++tot] = E(S, id1[i], 1);
G[++tot] = E(id2[i], T, 1);
for (int j = 1; j <= n; ++j) {
for (int _ = 0; _ < c[i][j]; ++_) {
G[++tot] = E(id1[i], id2[j], 1);
}
}
}
for (int p = 1; p <= m; ++p) {
len = 1;
mems(head, 0);
for (int i = 1; i <= tot; ++i) {
int u = G[i].u, v = G[i].v, d = G[i].d;
solver.add(u, v, d, i);
}
if (solver.solve() != n) {
puts("No");
return;
}
for (int i = head[S]; i; i = edges[i].next) {
edge e1 = edges[i];
if (!(i & 1) && e1.flow) {
for (int j = head[e1.to]; j; j = edges[j].next) {
edge e2 = edges[j];
if (!(j & 1) && e2.flow) {
int id = e2.id;
--G[id].d;
int x = e1.to, w = e2.to - n;
b[x][p] = w;
}
}
}
}
}
puts("Yes");
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= m; ++j) {
printf("%d ", b[i][j]);
}
putchar('\n');
}
}
int main() {
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}
