洛谷 P2065 [TJOI2011]卡片 网络流

题面

题目传送门

解法

显然的二分图最大匹配

但是，如果连接\(nm\)条边的话肯定会出事

考虑如何减少边的数量

将每一个数分解质因数，将所有出现过的质因数放在中间一排

对于每一个\(a_i\)和\(b_i\)，和中间一排能整除自己的质因数连边

这样就可以把边数降低到\(O(n)\)级别

然后dinic即可

代码

#include <bits/stdc++.h>
#define N 1010
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
	x = 0; int f = 1; char c = getchar();
	while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
	while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Edge {
	int next, num, c;
} e[N * N];
int s, t, cnt, a[N], b[N], l[N * 2], cur[N * 2], num[N * 2];
void add(int x, int y, int c) {
	e[++cnt] = (Edge) {e[x].next, y, c};
	e[x].next = cnt;
}
void Add(int x, int y, int c) {
	add(x, y, c), add(y, x, 0);
}
bool bfs(int s) {
	for (int i = 1; i <= t; i++) l[i] = -1;
	queue <int> q; q.push(s);
	while (!q.empty()) {
		int x = q.front(); q.pop();
		for (int p = e[x].next; p; p = e[p].next) {
			int k = e[p].num, c = e[p].c;
			if (c && l[k] == -1)
				l[k] = l[x] + 1, q.push(k);
		}
	}
	return l[t] != -1;
}
int dfs(int x, int lim) {
	if (x == t) return lim;
	int used = 0;
	for (int p = cur[x]; p; p = e[p].next) {
		int k = e[p].num, c = e[p].c;
		if (c && l[k] == l[x] + 1) {
			int w = dfs(k, min(c, lim - used));
			e[p].c -= w, e[p ^ 1].c += w;
			if (e[p].c) cur[x] = p; used += w;
			if (used == lim) return lim;
		}
	}
	if (!used) l[x] = -1;
	return used;
}
int dinic() {
	int ret = 0;
	while (bfs(s)) {
		for (int i = 0; i <= t; i++)
			cur[i] = e[i].next;
		ret += dfs(s, INT_MAX);
	}
	return ret;
}
int main() {
	int T; read(T);
	while (T--) {
		int n, m; read(n), read(m);
		set <int> st;
		for (int i = 1; i <= n; i++) {
			read(a[i]); int x = a[i];
			for (int j = 2; j * j <= x; j++) {
				if (x % j == 0) {
					st.insert(j);
					while (x % j == 0) x /= j;
				}
			}
			if (x > 1) st.insert(x);
		}
		for (int i = 1; i <= m; i++) {
			read(b[i]); int x = b[i];
			for (int j = 2; j * j <= x; j++) {
				if (x % j == 0) {
					st.insert(j);
					while (x % j == 0) x /= j;
				}
			}
			if (x > 1) st.insert(x);
		}
		int len = 0;
		for (set <int> :: iterator it = st.begin(); it != st.end(); it++)
			num[++len] = *it;
		s = 0, t = cnt = n + m + len + 1;
		if (cnt % 2 == 0) cnt++;
		for (int i = 0; i <= t; i++) e[i].next = 0;
		for (int i = 1; i <= n; i++) Add(s, i, 1);
		for (int i = 1; i <= m; i++) Add(i + n + len, t, 1);
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= len && num[j] <= a[i]; j++)
				if (a[i] % num[j] == 0) Add(i, j + n, 1);
		for (int i = 1; i <= m; i++)
			for (int j = 1; j <= len && num[j] <= b[i]; j++)
				if (b[i] % num[j] == 0) Add(j + n, i + len + n, 1);
		cout << dinic() << "\n";
	}
	return 0;
}