BZOJ3275 Number

偶数之间gcd不为1，显然

奇数^2加奇数^2不等于偶数^2，因为奇数可以写成2x+1的形式，所以奇数方模4余1，两个加一起余2，而偶数方模4余0

二分图，跑最小割即可

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<iomanip>
#include<cmath>
#include<cstring>
#include<ctime>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<bitset>
#include<map>
using namespace std;
#define MAXN 3010
#define MAXM 5000010
#define INF 1000000000000ll
#define MOD 1000000007
#define ll long long
#define eps 1e-8
struct vec{
int to;
int fro;
ll v;
};
vec mp[MAXM];
int tai[MAXN],cnt=1;
int s,t;
int d[MAXN];
int q[MAXN],hd,tl;
ll ans;
int cur[MAXN];
inline void be(int x,int y,ll z){
mp[++cnt].to=y;
mp[cnt].fro=tai[x];
tai[x]=cnt;
mp[cnt].v=z;
}
inline void bse(int x,int y,ll z){
be(x,y,z);
be(y,x,0);
}
bool bfs(){
int i,x,y;
memset(d,0,sizeof(d));
d[s]=1;
hd=tl=0;
q[tl++]=s;
while(hd!=tl){
x=q[hd++];
for(i=tai[x];i;i=mp[i].fro){
y=mp[i].to;
if(mp[i].v&&(!d[y])){
d[y]=d[x]+1;
q[tl++]=y;
}
}
}

return d[t];
}
ll dfs(int x,ll mx){
if(x==t){
return mx;
}
ll i,y,tt;
ll re=0;
for(i=tai[x];i;i=mp[i].fro){
y=mp[i].to;
if(d[y]==d[x]+1&&mp[i].v){
tt=dfs(y,min(mx,mp[i].v));
mp[i].v-=tt;
mp[i^1].v+=tt;
mx-=tt;
re+=tt;
if(!mx){
return re;
}
}
}
if(!re){
d[x]=0;
}
return re;
}
int n,m;
ll a[MAXN],b[MAXN];
ll gcd(ll x,ll y){
return !y?x:gcd(y,x%y);
}
int main(){
int i,j,x,y,k,xx,yy;
scanf("%d",&n);
for(i=1;i<=n;i++){
scanf("%lld",&a[i]);
}
s=n+1;
t=s+1;
for(i=1;i<=n;i++){
if(a[i]&1){
bse(s,i,a[i]);
}else{
bse(i,t,a[i]);
}
ans+=a[i];
for(j=1;j<=i;j++){
if((a[i]&1)==(a[j]&1)){
continue ;
}
if(gcd(a[i],a[j])==1&&((ll)sqrt(a[i]*a[i]+a[j]*a[j]))*((ll)sqrt(a[i]*a[i]+a[j]*a[j]))==a[i]*a[i]+a[j]*a[j]){
if(a[i]&1){
bse(i,j,INF);
}else{
bse(j,i,INF);
}
}
}
}
while(bfs()){
ans-=dfs(s,INF);
}
printf("%lld\n",ans);
return 0;
}

/*
2

1 2

3 5

*/