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AtCoder Beginner Contest 317 G Rearranging

发布时间 2023-08-27 21:46:43作者: zltzlt

洛谷传送门

AtCoder 传送门

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 \(|S| \ge |N(S)|\)。

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 \(m|S| \ge |N(S)|\)。

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;
}