【网络流】 codeforces 498C Array and Operations

找出所有的因子跑最大流即可。。。当然本题还可以用二分匹配做。。。

#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 2000005
#define eps 1e-10
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
//#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head


struct Edge
{
	int v, c, next;
	Edge() {}
	Edge(int v, int c, int next) : v(v), c(c), next(next) {}
}E[maxm];

queue<int> q;
int H[maxn], cntE;
int dis[maxn];
int cur[maxn];
int cnt[maxn];
int pre[maxn];
int flow, s, t, nv;
int n, m, o;
int a[maxn];
int x[maxn];
vector<pair<int, int> > g[maxn];

map<int, int> mpp;
map<int, int>::iterator it;

void addedges(int u, int v, int c)
{
	//printf("u = %d v = %d c = %d\n", u, v, c);
	E[cntE] = Edge(v, c, H[u]);
	H[u] = cntE++;
	E[cntE] = Edge(u, 0, H[v]);
	H[v] = cntE++;
}

void bfs(void)
{
	memset(cnt, 0, sizeof cnt);
	memset(dis, -1, sizeof dis);
	cnt[0] = 1, dis[t] = 0;
	q.push(t);
	while(!q.empty()) {
		int u = q.front();
		q.pop();
		for(int e = H[u]; ~e; e = E[e].next) {
			int v = E[e].v;
			if(dis[v] == -1) {
				dis[v] = dis[u] + 1;
				cnt[dis[v]]++;
				q.push(v);
			}
		}
	}
}

int isap(void)
{
	memcpy(cur, H, sizeof cur);
	flow = 0;
	bfs();
	int u = pre[s] = s, e, minv, f, pos;
	while(dis[s] < nv) {
		if(u == t) {
			f = INF;
			for(int i = s; i != t; i = E[cur[i]].v) if(E[cur[i]].c < f) {
				f = E[cur[i]].c;
				pos = i;
			}
			for(int i = s; i != t; i = E[cur[i]].v) {
				E[cur[i]].c -= f;
				E[cur[i] ^ 1].c += f;
			}
			flow += f;
			u = pos;
		}
		for(e = H[u]; ~e; e = E[e].next) if(E[e].c && dis[E[e].v] + 1 == dis[u]) break;
		if(~e) {
			cur[u] = e;
			pre[E[e].v] = u;
			u = E[e].v;
		}
		else {
			if(--cnt[dis[u]] == 0) break;
			for(e = H[u], minv = nv; ~e; e = E[e].next) if(E[e].c && minv > dis[E[e].v]) {
				minv = dis[E[e].v];
				cur[u] = e;
			}
			dis[u] = minv + 1;
			cnt[dis[u]]++;
			u = pre[u];
		}
	}
	return flow;
}

void init(void)
{
	cntE = 0;
	mpp.clear();
	memset(H, -1, sizeof H);
}

void read(void)
{
	int aa, bb;
	scanf("%d%d", &n, &m);
	s = 0, t = n + 1, o = n + 2;
	for(int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
		for(int j = 2; (LL)j * j <= a[i]; j++) {
			if(a[i] % j == 0) {
				int t = 0;
				while(a[i] % j == 0) a[i] /= j, t++;
				g[i].push_back(mp(j, t));
			}
		}
		if(a[i] != 1) g[i].push_back(mp(a[i], 1));
	}
	for(int i = 1; i <= n; i++) x[i] = g[i].size();
	for(int i = 1; i <= n; i++) x[i] += x[i-1];
	while(m--) {
		scanf("%d%d", &aa, &bb);
		if(aa % 2 == 0) swap(aa, bb);
		for(int i = 0; i < g[aa].size(); i++)
			for(int j = 0; j < g[bb].size(); j++)
				if(g[aa][i].first == g[bb][j].first)
					addedges(x[aa-1] + i + o, x[bb-1] + j + o, INF);
	}
	for(int i = 1; i <= n; i++)
		if(i % 2) {
			for(int j = 0; j < g[i].size(); j++)
				addedges(s, x[i-1] + j + o, g[i][j].second);
		}
		else {
			for(int j = 0; j < g[i].size(); j++)
				addedges(x[i-1] + j + o, t, g[i][j].second);
		}
	nv = x[n] + o + 1;
}

void work(void)
{
	printf("%d\n", isap());
}

int main(void)
{
	init();
	read();
	work();

	return 0;
}