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題目傳送門

解法

顯然的二分圖最大匹配

但是，如果連接\(nm\)條邊的話肯定會出事

考慮如何減少邊的數量

將每一個數分解質因數，將所有出現過的質因數放在中間一排

對於每一個\(a_i\)和\(b_i\)，和中間一排能整除自己的質因數連邊

這樣就可以把邊數降低到\(O(n)\)級別

然後dinic即可

代碼

#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;
}
 

洛谷 P2065 [TJOI2011]卡片 網絡流