Luogu2831

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EndSaH的代码

#include<bits/stdc++.h>using namespace std;const int MAXN=20;
const double eps=1e-6;struct Point{double x,y;
}p[MAXN],surp[MAXN];struct Para{double a,b;
}par[MAXN];int T,n,m,ans;inline bool equal(double x,double y){return x-y>=-eps&&x-y<=eps;
}inline bool inc(double a,double b,double x,double y){return equal(a*x*x+b*x,y);
}inline void build(double &a,double &b,double x1,double y1,double x2,double y2){a=(x2*y1-x1*y2)/(x1*x2*(x1-x2));b=(x1*x1*y2-x2*x2*y1)/(x1*x2*(x1-x2));//矩阵乘法+逆矩阵求法
}inline void dfs(int now,int num,int rest){//DFS最重要的是分清楚情况 if(num+rest>=ans) return;if(now>n){ans=num+rest;return;}for(int i=1;i<=num;i++){if(inc(par[i].a,par[i].b,p[now].x,p[now].y)){dfs(now+1,num,rest);//此点已被覆盖 return;}}for(int i=1;i<=rest;i++){if(equal(p[i].x,p[now].x)) continue;double a,b;build(a,b,surp[i].x,surp[i].y,p[now].x,p[now].y);//组成新抛物线 if(a<0){par[num+1]=(Para){a,b};Point tmp=surp[i];for(int j=i;j<=rest-1;j++)surp[j]=surp[j+1];dfs(now+1,num+1,rest-1);for(int j=rest;j>=i+1;j--)surp[j]=surp[j-1];surp[i]=tmp;}}surp[rest+1]=p[now];//不与任何点组成新抛物线 dfs(now+1,num,rest+1);
}int main(){scanf("%d",&T);while(T--){ans=1e9;scanf("%d%d",&n,&m);for(int i=1;i<=n;i++){scanf("%lf%lf",&p[i].x,&p[i].y);}dfs(1,0,0);printf("%d\n",ans);}
}

\(DP\)做法

#include<bits/stdc++.h>
using namespace std;
const int INF=1e8;
double x[20],y[20];
int para[200],dp[1<<18];
int n,m,countpara;
int Min(int x,int y){return x<y?x:y;}
void build(double &a,double &b,double x1,double y1,double x2,double y2){a=(x2*y1-x1*y2)/(x1*x2*(x1-x2));b=(x1*x1*y2-x2*x2*y1)/(x1*x2*(x1-x2));//矩阵乘法+逆矩阵求法
}
bool inc(double a,double b,double x,double y){double t=a*x*x+b*x-y;return t>=-1e-6&&t<=1e-6;//精度判断
}
void pre(){scanf("%d%d",&n,&m);for(int i=0;i<(1<<n);i++)dp[i]=INF;countpara=0;    for(int i=0;i<n;i++)scanf("%lf%lf",&x[i],&y[i]);for(int i=0;i<n;i++){para[countpara++]=(1<<i);for(int j=i+1,vis=0;j<n;j++){if(vis&(1<<j)) continue;double a,b;build(a,b,x[i],y[i],x[j],y[j]);if(a>=0) continue;para[countpara]=(1<<i);for(int k=j;k<n;k++){if(inc(a,b,x[k],y[k])){vis|=(1<<k);para[countpara]|=(1<<k);//这一条有效抛物线的状态 }}countpara++;}}
}
int ans(){dp[0]=0;for(int i=0;i<(1<<n);i++)//从0开始 for(int j=0;j<countpara;j++)//dp[para[j]]=0;dp[i|para[j]]=Min(dp[i|para[j]],dp[i]+1);//逐渐更新出最优解 return dp[(1<<n)-1];
}
int main(){int T;scanf("%d",&T);while(T--){pre();printf("%d\n",ans());}
}