[ABC263G] Erasing Prime Pairs
Problem Statement
There are integers with $N$ different values written on a blackboard. The $i$-th value is $A_i$ and is written $B_i$ times.
You may repeat the following operation as many times as possible:
- Choose two integers $x$ and $y$ written on the blackboard such that $x+y$ is prime. Erase these two integers.
Find the maximum number of times the operation can be performed.
Constraints
- $1 \leq N \leq 100$
- $1 \leq A_i \leq 10^7$
- $1 \leq B_i \leq 10^9$
- All $A_i$ are distinct.
- All values in input are integers.
Input
Input is given from Standard Input in the following format:
$N$ $A_1$ $B_1$ $A_2$ $B_2$ $\vdots$ $A_N$ $B_N$
Output
Print the answer.
Sample Input 1
3 3 3 2 4 6 2
Sample Output 1
3
We have $2 + 3 = 5$, and $5$ is prime, so you can choose $2$ and $3$ to erase them, but nothing else. Since there are four $2$s and three $3$s, you can do the operation three times.
Sample Input 2
1 1 4
Sample Output 2
2
We have $1 + 1 = 2$, and $2$ is prime, so you can choose $1$ and $1$ to erase them. Since there are four $1$s, you can do the operation twice.
既然要判断质数，一个肯定要做的操作就是把 \(2\times10^7\) 之内的质数筛出来。不妨把和为质数的两个点连条边。边相连的两个点的点权可以同时减去一个数，要保证点权为正数，问最多操作次数。
手动模拟一下，有网络流的感觉。于是考虑最大流解决。把第 \(i\) 种数拆成两个点，第一个点编号为 \(i\)，第二个点编号为 \(n+i\)。对于第 \(i\) 个数，从源点 \(s\) 连一条流量为 \(b_i\) 的边，从 \(i+n\) 连一条流量为 \(b_i\) 的边至汇点 \(t\)。如果 \(a_i+a_j\) 为质数，从 \(i\) 连一条流量为 \(+\infty\) 的边至 \(j+n\)。然后跑一次最大流，得出答案后除以 \(2\) 就是答案。
为什么要除以 \(2\)？由于我们计算最大流时，\(a_i+a_j\) 若为质数，那么 \(i\) 到 \(j+n\) 的边中算了一次，\(j+n\) 至 \(i\) 的边中算了一次。所以算了两次。
但是我们没考虑 \(1+1=2\)？其实计算最大流时已经计算。须要两个 \(1\) 才能有一次删除机会，而我们最后也除以了 \(2\)。所以不需要特殊判断。
#include<bits/stdc++.h>
using namespace std;
const int N=205,M=40005,P=2e7+5;
int n,m,s,t,k,g,hd[N],l,r,v[N],q[N],vhd[N],uu,vv,ww,e_num=1,p[P],p_num,pri[P],a[N],b[N],ee;
struct edge{
int v,nxt,w;
}e[M<<1];
void add_edge(int u,int v,int w)
{
e[++e_num]=(edge){v,hd[u],w};
hd[u]=e_num;
}
long long ans;
int bfs()
{
memset(v,0,sizeof(v));
memcpy(hd,vhd,sizeof(hd));
q[l=r=1]=s,v[s]=1;
while(l<=r)
{
if(q[l]==t)
break;
for(int i=hd[q[l]];i;i=e[i].nxt)
if(e[i].w>0&&!v[e[i].v])
q[++r]=e[i].v,v[e[i].v]=v[q[l]]+1;
++l;
}
return v[t];
}
int dfs(int x,int s)
{
// printf("%d %d\n",x,s);
if(x==t)
return s;
int b,c;
for(int&i=hd[x];i;i=e[i].nxt)
{
b=e[i].v,c=min(s,e[i].w);
if(v[b]==v[x]+1&&c>0)
{
g=dfs(b,c);
if(g)
{
e[i].w-=g;
e[i^1].w+=g;
return g;
}
else
v[b]=0;
}
}
return 0;
}
signed main()
{
scanf("%d",&n);
s=0,t=2*n+1;
for(int i=2;i<P;i++)
{
if(!pri[i])
p[++p_num]=i;
for(int j=1;j<=p_num&&1LL*p[j]*i<P;j++)
{
pri[p[j]*i]=1;
if(i%p[j]==0)
break;
}
}
for(int i=1;i<=n;i++)
scanf("%lld%lld",a+i,b+i);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(!pri[a[i]+a[j]])
{
add_edge(i,n+j,2e9);
add_edge(j+n,i,0);
}
}
}
for(int i=1;i<=n;i++)
{
add_edge(s,i,b[i]);
add_edge(i,s,0);
}
for(int i=1;i<=n;i++)
{
add_edge(i+n,t,b[i]);
add_edge(t,i+n,0);
}
memcpy(vhd,hd,sizeof(hd));
while(bfs())
{
while(k=dfs(s,2e9))
ans+=k;
}
printf("%lld",ans/2);
return 0;
}
