luogu P2765 魔术球问题 (最小路径覆盖)

大意:给定n根柱子, 依次放入1,2,3,...的球, 同一根柱子相邻两个球和为完全平方数, 求最多放多少个球.

对和为平方数的点连边, 就相当于求DAG上最小路径覆盖.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, P2 = 998244353, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head

const int N = 1e6+10;
int n, clk, deg[N], nxt[N];
int c[N], fa[N], vis[N];
vector<int> g[N];
int dfs(int x) {
	for (int y:g[x]) if (vis[y]!=clk) {
		vis[y] = clk;
		if (!fa[y]||dfs(fa[y])) return fa[y]=x;
	}
	return 0;
}
int chk(int x) {
	REP(i,1,2*x) g[i].clear(),fa[i]=0;
	REP(i,1,x) REP(j,i+1,x) if (c[i+j]) {
		g[i].pb(j+x),g[j+x].pb(i);
	}
	int ans = 0;
	REP(i,1,x) ++clk, ans+=!!dfs(i);
	return x-ans<=n;
}

int main() {
	REP(i,1,1000) c[i*i]=1;
	scanf("%d", &n);
	int ans = 1;
	while (chk(ans)) ++ans;
	chk(--ans);
	printf("%d\n", ans);
	REP(i,1,ans) if (fa[i]) nxt[i]=fa[i]-ans,deg[fa[i]-ans]=1;
	REP(i,ans+1,2*ans) if (fa[i]) nxt[fa[i]]=i-ans,deg[i-ans]=1;
	REP(i,1,ans) if (!deg[i]) {
		int x = i;
		do printf("%d ",x),x=nxt[x]; while (x);
		hr;
	}
}