网络流（最大流）：CodeForces 499E Array and Operations

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

In one operation you can perform a sequence of actions:

take one of the good pairs (i_k, j_k) and some integer v (v > 1), which divides both numbers a[i_k] and a[j_k];
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[n] (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 i_k, j_k (1 ≤ i_k < j_k ≤ n, i_k + j_k is an odd number).

It is guaranteed that all the good pairs are distinct.

Output

Output the answer for the problem.

Sample Input

Input

3 28 3 81 22 3

Output

Input

3 28 12 81 22 3

Output

2　　将点拆成多个素数，然后套用网络流，还有一种思路是建很多次图，分开处理素数的匹配。

  1 #include <iostream>
  2 #include <cstring>
  3 #include <cstdio>
  4 using namespace std;
  5 const int INF=1000000000;
  6 const int N=110,M=1010,K=35010;
  7 int P[K],cntx;bool check[K];
  8 void Linear_Shaker(){
  9     for(int i=2;i<K;i++){
 10         if(!check[i])P[++cntx]=i;
 11         for(int j=1;j<=cntx;j++){
 12             if(i*P[j]>=K)break;
 13             check[i*P[j]]=true;
 14             if(i%P[j]==0)break;
 15         }
 16     }
 17 }
 18 int v[N][12],c[N][12],id[N][12],h[N],idx;
 19 int cnt=1,fir[M],to[M*20],nxt[M*20],cap[M*20];
 20 int dis[M],gap[M],path[M],fron[M],q[M],f,b;
 21 void add(int a,int b,int c){
 22     nxt[++cnt]=fir[a];
 23     to[fir[a]=cnt]=b;
 24     cap[cnt]=c;
 25 }
 26
 27 void addedge(int a,int b,int c){
 28     //printf("%d %d\n",a,b);
 29     add(a,b,c);add(b,a,0);
 30 }
 31
 32 bool BFS(int S,int T){
 33     q[f=b]=T;dis[T]=1;
 34     while(f<=b){
 35         int x=q[f++];
 36         for(int i=fir[x];i;i=nxt[i])
 37             if(!dis[to[i]]){
 38                 dis[to[i]]=dis[x]+1;
 39                 q[++b]=to[i];
 40             }
 41     }
 42     return dis[S];
 43 }
 44
 45 int Max_Flow(int S,int T){
 46     if(!BFS(S,T))return 0;
 47     for(int i=S;i<=T;i++)fron[i]=fir[i];
 48     for(int i=S;i<=T;i++)gap[dis[i]]+=1;
 49     int ret=0,f,p=S;
 50     while(dis[S]<=T+1){
 51         if(p==T){
 52             f=INF;
 53             while(p!=S){
 54                 f=min(f,cap[path[p]]);
 55                 p=to[path[p]^1];
 56             }ret+=f,p=T;
 57             while(p!=S){
 58                 cap[path[p]]-=f;
 59                 cap[path[p]^1]+=f;
 60                 p=to[path[p]^1];
 61             }
 62         }
 63         for(int &i=fron[p];i;i=nxt[i])
 64             if(cap[i]&&dis[to[i]]==dis[p]-1){
 65                 path[p=to[i]]=i;break;
 66             }
 67         if(!fron[p]){
 68             if(!--gap[dis[p]])break;int Min=T+1;
 69             for(int i=fir[p];i;i=nxt[i])
 70                 if(cap[i])Min=min(Min,dis[to[i]]);
 71             gap[dis[p]=Min+1]+=1;fron[p]=fir[p];
 72             if(p!=S)p=to[path[p]^1];
 73         }
 74     }
 75     return ret;
 76 }
 77
 78 int n,m,S,T,num[N];
 79 int G[N][N];
 80 int main(){
 81     Linear_Shaker();
 82     scanf("%d%d",&n,&m);
 83     for(int i=1;i<=n;i++){
 84         scanf("%d",&num[i]);
 85         for(int j=1;j<=cntx;j++){
 86             if(P[j]*P[j]>num[i])break;
 87             if(num[i]%P[j]==0){
 88                 v[i][h[i]]=P[j];
 89                 while(num[i]%P[j]==0){
 90                     c[i][h[i]]+=1;
 91                     num[i]/=P[j];
 92                 }h[i]+=1;
 93             }
 94         }
 95         if(num[i]!=1){
 96             v[i][h[i]]=num[i];
 97             c[i][h[i]++]=1;
 98         }
 99     }
100     /*
101     for(int i=1;i<=n;i++){
102         for(int j=0;j<h[i];j++)
103             printf("<%d %d> ",v[i][j],c[i][j]);
104         puts("");
105     }
106     */
107     for(int i=1,a,b;i<=m;i++){
108         scanf("%d%d",&a,&b);
109         if(a%2)swap(a,b);
110         G[a][b]=1;
111     }
112     for(int i=1;i<=n;i++)
113         for(int j=0;j<h[i];j++)
114             id[i][j]=++idx;
115     S=0;T=idx+1;
116     for(int i=1;i<=n;i++)
117         for(int j=0;j<h[i];j++){
118             if(i%2==0)addedge(S,id[i][j],c[i][j]);
119             else addedge(id[i][j],T,c[i][j]);
120         }
121     for(int i=2;i<=n;i+=2)
122         for(int j=1;j<=n;j+=2)if(G[i][j])
123             for(int x=0;x<h[i];x++)
124                 for(int y=0;y<h[j];y++)
125                     if(v[i][x]==v[j][y])
126                         addedge(id[i][x],id[j][y],INF);
127     printf("%d\n",Max_Flow(S,T));
128     return 0;
129 }

转载于:https://www.cnblogs.com/TenderRun/p/5928206.html