abc282E - Choose Two and Eat One（转化成生成树）

E - Choose Two and Eat One
非常巧妙的一集

可以将整个局面看作一张图，选两个数获得的score就是它们的边权，然后做最大生成树，不难发现操作和建树之间是一一对应的。

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#define A puts("YES")
#define B puts("NO")
#define fo(i,a,b) for (int (i)=(a);(i)<=(b);(i)++)
#define fd(i,b,a) for (int (i)=(b);(i)>=(a);(i)--)
#define mk(x,y) make_pair((x),(y))
using namespace std;
typedef double db;
typedef long long ll;
const int N=500+5;
const ll inf=1ll<<60;
struct node{
	ll x,y,v;
};
node v[N*N];

ll f[N],a[N],n,m,tot,f1,f2;
ll find(ll x){
	if (x==f[x]) return x;
	return f[x]=find(f[x]);
}
ll power(ll a,ll b){
	ll t=1,y=a;
	while (b){
		if (b&1) t=t*y%m;
		y=y*y%m;
		b/=2;
	}
	return t;
}
bool cmp(node a,node b){
	return a.v>b.v;
}
int main()
{ 
//	freopen("data.in","r",stdin);
	
	scanf("%lld %lld",&n,&m);
	fo(i,1,n) scanf("%lld",&a[i]);
	
	fo(i,1,n) {
		fo(j,i+1,n) {
			v[++tot]=(node){i,j, (power(a[i],a[j])+power(a[j],a[i]))%m };
		}
	}
	
	sort(v+1,v+tot+1,cmp);
	
	fo(i,1,n) f[i]=i;
	
	ll ans=0;
	fo(i,1,tot) {
		f1=find(v[i].x);
		f2=find(v[i].y);
		if (f1^f2) {
			ans+=v[i].v;
			f[f1]=f2;
		}
	}
	
	printf("%lld",ans);
	return 0;
}