CF498C. Array and Operations [二分图]

CF498C. Array and Operations

题意：

给定一个长为 n 的数组，以及 m 对下标 (a, b) 且满足 a + b 为奇数，每次操作可以将同一组的两个数同时除以一个公约数

问最多能进行多少次操作

\[1≤n,m ≤100,1≤ai ≤10^9 \]

---

根据奇偶性二分图定理此题必定考二分图

贪心，每次除一个质数

质数之间是独立的，可以分开考虑每一个质因子

建图：s -x中质因子p数量-> x -inf-> y -y中质因子p数量-> t

最大权匹配就是这个质因子能带来的最多操作数

注意质因子分解别写错了，最后判x>1

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
#include <set>
#include <map>

#define fir first
#define sec second
using namespace std;
const int N = 105, inf = 1e9;


int n, m, a[N];
pair<int, int> b[N];
map<int, int> li[N];
set<int> s;

void fac(int p) { 
	int x = a[p], sx = sqrt(x) + 1; //printf("fac %d\n", a[p]);
	for(int i=2; i<=sx; i++) if(x % i == 0) { //printf("iii %d\n", i);
		int cnt = 0;
		while(x%i == 0) cnt++, x/=i; //printf("x %d\n", x);
		//li[p].push_back(make_pair(i, cnt));
		li[p][i] = cnt;
		s.insert(i);
	}
	if(x > 1) li[p][x] = 1, s.insert(x);
}

namespace mf {
	int s, t;
	struct edge {int v, ne, c, f;} e[1005];
	int cnt=1, h[N];
	void ins(int u, int v, int c) { //printf("ins %d %d %d\n", u, v, c);
		e[++cnt] = (edge) {v, h[u], c, 0}; h[u] = cnt;
		e[++cnt] = (edge) {u, h[v], 0, 0}; h[v] = cnt;
	}
	int cur[N], vis[N], d[N], head, tail, q[N];
	bool bfs() {
		memset(vis, 0, sizeof(vis));
		head = tail = 1;
		q[tail++] = s; d[s] = 0; vis[s] = 1;
		while(head != tail) {
			int u = q[head++];
			for(int i=h[u]; i; i=e[i].ne) {
				int v = e[i].v;
				if(!vis[v] && e[i].c > e[i].f) {
					vis[v] = 1;
					d[v] = d[u] + 1;
					q[tail++] = v;
					if(v == t) return true;
				}
			}
		}
		return false;
	}
	int dfs(int u, int a) {
		if(u==t || a==0) return a;
		int flow = 0, f;
		for(int &i=cur[u]; i; i=e[i].ne) {
			int v = e[i].v;
			if(d[v] == d[u]+1 && (f = dfs(v, min(a, e[i].c-e[i].f))) > 0) {
				flow += f;
				e[i].f += f;
				e[i^1].f -= f;
				a -= f;
				if(a==0) break;
			}
		}
		if(a) d[u] = -1;
		return flow;
	}

	void build(int p) {
		cnt = 1;
		memset(h, 0, sizeof(h));
		s=0; t=n+1;
		for(int i=1; i<=m; i++) ins(b[i].fir, b[i].sec, inf);
		for(int i=1; i<=n; i++) if(li[i].count(p)) {
			if(i & 1) ins(s, i, li[i][p]);
			else ins(i, t, li[i][p]);
		}
	}

	int solve(int p) { //printf("\nsolve %d\n", p);
		build(p);
		int flow = 0;
		while(bfs()) {
			for(int i=s; i<=t; i++) cur[i] = h[i];
			flow += dfs(s, inf);
		}
		return flow;
	}

}
int main() {
	//freopen("in", "r", stdin);
	ios::sync_with_stdio(false); cin.tie(); cout.tie();

	cin >> n >> m;
	for(int i=1; i<=n; i++) cin >> a[i];
	for(int i=1; i<=m; i++) {
		cin >> b[i].fir >> b[i].sec;
		if(b[i].sec & 1) swap(b[i].fir, b[i].sec);
	}

	for(int i=1; i<=n; i++) fac(i);
	int ans = 0;
	
	for(set<int>::iterator it = s.begin(); it != s.end(); it++) ans += mf::solve(*it);
	cout << ans;
}