传送门
每个点拆点xi,yi,s->xi,yi->t,bi
对于不能同时选的两个点ij,xi->yj,inf,xj->yi,inf
答案为 (2∗∑i=1nbi−maxflow)/2
刚开始的时候只对于j>i的连边,然后 ∑i=1nbi−maxflow ,但是wa了,不是很理解为什么…感觉这两种做法是等价的啊…
可能是因为图不是对称的所以会出现割一边的情况?
#include
#include
#include
#include
#include
#include
using namespace std;
#define LL long long
#define N 2005
#define inf 2000000000
int n,b,s,t,sum,maxflow;
int tot,point[N],nxt[N*N],v[N*N],remain[N*N];
int deep[N],last[N],num[N],cur[N];
LL a[N];
queue <int> q;
void addedge(int x,int y,int cap)
{
++tot; nxt[tot]=point[x]; point[x]=tot; v[tot]=y; remain[tot]=cap;
++tot; nxt[tot]=point[y]; point[y]=tot; v[tot]=x; remain[tot]=0;
}
void bfs(int t)
{
for (int i=s;i<=t;++i) deep[i]=t;
deep[t]=0;
for (int i=s;i<=t;++i) cur[i]=point[i];
while (!q.empty()) q.pop();
q.push(t);
while (!q.empty())
{
int now=q.front();q.pop();
for (int i=point[now];i!=-1;i=nxt[i])
if (deep[v[i]]==t&&remain[i^1])
{
deep[v[i]]=deep[now]+1;
q.push(v[i]);
}
}
}
int addflow(int s,int t)
{
int now=t,ans=inf;
while (now!=s)
{
ans=min(ans,remain[last[now]]);
now=v[last[now]^1];
}
now=t;
while (now!=s)
{
remain[last[now]]-=ans;
remain[last[now]^1]+=ans;
now=v[last[now]^1];
}
return ans;
}
void isap(int s,int t)
{
bfs(t);
for (int i=s;i<=t;++i) ++num[deep[i]];
int now=s;
while (deep[s]if (now==t)
{
maxflow+=addflow(s,t);
now=s;
}
bool has_find=false;
for (int i=cur[now];i!=-1;i=nxt[i])
if (deep[v[i]]+1==deep[now]&&remain[i])
{
has_find=true;
cur[now]=i;
last[v[i]]=i;
now=v[i];
break;
}
if (!has_find)
{
int minn=t-1;
for (int i=point[now];i!=-1;i=nxt[i])
if (remain[i]) minn=min(minn,deep[v[i]]);
if (!(--num[deep[now]])) break;
++num[deep[now]=minn+1];
cur[now]=point[now];
if (now!=s) now=v[last[now]^1];
}
}
}
bool check1(LL a,LL b)
{
LL T=a*a+b*b;
LL sT=sqrt(T);
if (sT*sT==T) return 0;
else return 1;
}
LL gcd(LL a,LL b)
{
if (!b) return a;
else return gcd(b,a%b);
}
bool check2(LL a,LL b)
{
LL T=gcd(a,b);
if (T>1) return 1;
else return 0;
}
int main()
{
tot=-1;memset(point,-1,sizeof(point));
scanf("%d",&n);
s=1,t=n+n+2;
for (int i=1;i<=n;++i) scanf("%lld",&a[i]);
for (int i=1;i<=n;++i)
{
scanf("%d",&b);
addedge(s,1+i,b);
addedge(1+n+i,t,b);
sum+=b*2;
}
for (int i=1;i<=n;++i)
for (int j=1;j<=n;++j)
{
bool flag=check1(a[i],a[j])||check2(a[i],a[j]);
if (!flag) addedge(1+i,1+n+j,inf);
}
isap(s,t);
printf("%d\n",(sum-maxflow)/2);
}