「CF498C」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), …, (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.
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 ik, jk (1 ≤ ik < jk ≤ n, ik + jk is an odd number).
It is guaranteed that all the good pairs are distinct.
Output the answer for the problem.
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1 2 3 4 |
3 2 8 3 8 1 2 2 3 |
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1 |
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1 2 3 4 |
3 2 8 12 8 1 2 2 3 |
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1 |
2 |
题解
题目显然给出的是二分图
每个质因数分开考虑
将有关的点连inf的边，源向奇点连容量为因子数量的边，偶点向汇连
作最大流
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#include<cstdio> #include<cmath> #include<ctime> #include<cstring> #include<iostream> #include<algorithm> #include<queue> #include<set> #define ll long long #define mod 1000000007 #define inf 1000000000 using namespace std; int read() { int f=1,x=0;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } int n,m,T,cnt,ans; int a[205],u[205],v[205]; int h[205],q[205],last[205],cur[205]; struct edge{ int to,next,v; }e[200005]; void insert(int u,int v,int w) { e[++cnt].to=v;e[cnt].next=last[u];last[u]=cnt;e[cnt].v=w; e[++cnt].to=u;e[cnt].next=last[v];last[v]=cnt;e[cnt].v=0; }; bool bfs() { int head=0,tail=1; for(int i=0;i<=T;i++)h[i]=-1; q[0]=0;h[0]=0; while(head!=tail) { int now=q[head];head++; for(int i=last[now];i;i=e[i].next) if(h[e[i].to]==-1&&e[i].v) { h[e[i].to]=h[now]+1; q[tail++]=e[i].to; } } return h[T]!=-1; } int dfs(int x,int f) { if(x==T)return f; int w,used=0; for(int i=cur[x];i;i=e[i].next) if(h[e[i].to]==h[x]+1) { w=dfs(e[i].to,min(e[i].v,f-used)); e[i].v-=w;e[i^1].v+=w; if(e[i].v)cur[x]=i; used+=w;if(used==f)return f; } if(!used)h[x]=-1; return used; } void dinic() { while(bfs()) { for(int i=0;i<=T;i++) cur[i]=last[i]; ans+=dfs(0,inf); } } void solve(int x) { cnt=1; memset(last,0,sizeof(last)); for(int i=1;i<=n;i++) { int t=0; while(a[i]%x==0)a[i]/=x,t++; if(i&1)insert(0,i,t); else insert(i+n,T,t); } for(int i=1;i<=m;i++) insert(u[i],v[i]+n,inf); dinic(); } int main() { n=read();m=read();T=2*n+1; for(int i=1;i<=n;i++)a[i]=read(); for(int i=1;i<=m;i++) { u[i]=read(),v[i]=read(); if(v[i]&1)swap(u[i],v[i]); } for(int k=1;k<=n;k++) { int t=sqrt(a[k]); for(int i=2;i<=t;i++) if(a[k]%i==0)solve(i); if(a[k]!=1)solve(a[k]); } printf("%d\n",ans); return 0; } |
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