Codeforces498C解题报告

C. Array and Operations

time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

You have written on a piece of paper an array of n positive integers a[1], a[2], …, a
and m good pairs of integers (i1, j1), (i2, j2), …, (im, jm). Each good pair (ik, jk) meets the following conditions: ik + jk is an odd number and 1 ≤ ik < jk ≤ n.

In one operation you can perform a sequence of actions:

take one of the good pairs (ik, jk) and some integer v (v > 1), which divides both numbers a[ik] and a[jk];

divide both numbers by v, i. e. perform the assignments: and .

Determine the maximum number of operations you can sequentially perform on the given array. Note that one pair may be used several times in the described operations.

Input

The first line contains two space-separated integers n, m (2 ≤ n ≤ 100, 1 ≤ m ≤ 100).

The second line contains n space-separated integers a[1], a[2], …, a
(1 ≤ a[i] ≤ 109) — the description of the array.

The following m lines contain the description of good pairs. The k-th line contains two space-separated integers ik, jk (1 ≤ ik < jk ≤ n, ik + jk is an odd number).

It is guaranteed that all the good pairs are distinct.

Output

Output the answer for the problem.

Sample test(s)

input

3 2

8 3 8

1 2

2 3

output

0

input

3 2

8 12 8

1 2

2 3

output

2

Solution：

考虑如何建图，由于“好”点对的和为一个奇数，不难想到(ui,vi)中ui与vi分别是一个奇数和一个偶数，然后建成二分图：暴力枚举因子数，相邻的u，v连无容量限制的边，从原点向奇数点集、偶数点集向汇点连出因子数的边，然后最大流为题目的解。

Code：

#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
#define maxn 200 + 10
#define maxm 40000 + 10
#define INF 0x7f7f7f7f
int n,m,s = 0,t, ans = 0,cur = 1, front[maxn], que[maxn], h[maxn], a[maxn], u[maxn], v[maxn];
struct edge{
int to, next, cap;
}e[maxm];
inline void addedge(int u, int v, int c){
cur ++;
e[cur].to = v;
e[cur].next = front[u];
e[cur].cap = c;
front[u] = cur;
cur ++;
e[cur].to = u;
e[cur].next = front[v];
e[cur].cap = 0;
front[v] = cur;
}
inline bool bfs(){
int x, tp = 0, d = 1;
memset(h, -1, sizeof(h));
que[tp] = h[0] = 0;
while(tp < d){
x = que[tp]; tp ++;
int now = front[x];
while(now){
if(e[now].cap && h[e[now].to] < 0){
que[d ++] = e[now].to;
h[e[now].to] = h[x] + 1;
}
now = e[now].next;
}
}
return h[t] == -1 ? 0 : 1;
}
inline int dfs(int x, int min_adv){
if(x == t) return min_adv;
int now = front[x];
int flow = 0, f;
while(now){
if(e[now].cap && h[e[now].to] == h[x] + 1){
f = min_adv - flow;
f = dfs(e[now].to, min(f, e[now].cap));
e[now].cap -= f;
e[now^1].cap += f;
flow += f;
if(flow == min_adv) return min_adv;
}
now = e[now].next;
}
if(!flow) h[x] = -1;
return flow;
}
inline void dinic(){
while(bfs()) ans += dfs(0, INF);
}
inline void calc(int x){
cur = 1; memset(front, 0, sizeof(front));
for(int i = 1; i <= n; i ++){
int tot = 0;
while(!(a[i] % x)) a[i] /= x, tot ++;
if(i & 1) addedge(s, i, tot);
else addedge(n + i, t, tot);
}
for(int i = 1; i <= m; i ++)addedge(u[i], n + v[i], INF);
dinic();
}
int main(){
scanf("%d%d",&n,&m); t = n << 1 | 1;
for(int i = 1; i <= n; i ++) scanf("%d",&a[i]);
for(int i = 1; i <= m; i ++){
scanf("%d%d",&u[i], &v[i]);
if(v[i] & 1) swap(u[i],v[i]);
}
for(int i = 1; i <= n; i ++){
for(int j = 2; j <= sqrt(a[i]); j ++)
if(!(a[i] % j)) calc(j);
if(a[i] != 1) calc(a[i]);
}
printf("%d\n",ans);
return 0;
}