P2831 愤怒的小鸟

Posted 2020-11-30 qyj060604

P2831 愤怒的小鸟

神仙状压dp

首先解一个非常休闲的二元一次方程组

接下来对于每两只鸟，我们计算出它们构成的二次函数（需判断合不合法）

接下来

记录函数上有几个点

然后开始愉快地dp

代码:

#include<bits/stdc++.h>
using namespace std;
typedef long double ld;
const int N=19;
int t;
int n,ppppp;
ld a[N],b[N];
bool clean[1<<N]={0};
bool yuy[1<<N]={0};
int f[1<<N];
int qu[1<<N]={0};
ld lpabs(ld x){
    if(x<0.0000001) return -x;
    return x;
}
bool cpair(ld x,ld y){
    if(lpabs(x-y)<0.0000001) return 1;
    return 0; 
} 
ld calx(ld x1,ld y1,ld x2,ld y2){
    return (y1*x2-x1*y2)/(x1*x1*x2-x1*x2*x2);
}
ld caly(ld x1,ld y1,ld x2,ld y2){
    return (x1*x1*y2-y1*x2*x2)/(x1*x1*x2-x1*x2*x2);
}
bool is_on(ld a,ld b,ld x,ld y){
    if(cpair(a*x*x+b*x,y)) return 1;
    return 0;
}
int main(){
    memset(f,127,sizeof(f));
    scanf("%d",&t);
    while(t--){
        memset(clean,0,sizeof(clean));
        memset(yuy,0,sizeof(yuy));
        memset(f,127,sizeof(f));
        memset(qu,0,sizeof(qu));
        scanf("%d%d",&n,&ppppp);
        for(int i=1;i<=n;i++){
            cin>>a[i]>>b[i];
        }
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(i==j) continue;
                if(cpair(a[i],a[j])) continue;
                ld dx,dy;
                dx=calx(a[i],b[i],a[j],b[j]);
                dy=caly(a[i],b[i],a[j],b[j]);
                //if(cpair(a[i]*a[i]*a[j]-a[i]*a[j]*a[j],0)||cpair(a[i]*a[i]*a[j]-a[i]*a[j]*a[j],0)) continue;
                if(dx>=0) continue;
                int cnt=0;
                int state=0;
                for(int k=1;k<=n;k++){
                    if(is_on(dx,dy,a[k],b[k])){
                        state+=(1<<(k-1));
                    }
                }
                for(int k=1;k<=n;k++){
                    if(is_on(dx,dy,a[k],b[k])){
                        for(int l=1;l<=n;l++){
                            if(is_on(dx,dy,a[l],b[l])){
                            //    cout<<k<<" -> "<<l<<endl;
                                clean[(1<<(k-1))|(1<<(l-1))]=1;
                                qu[(1<<(k-1))|(1<<(l-1))]=state;
                            }
                        }
                    }
                }
            }
        }
        f[0]=0;
        for(int i=1;i<=n;i++) yuy[(1<<(i-1))]=1;
        for(int i=1;i<=n;i++) f[1<<(i-1)]=1;
        for(int i=0;i<=(1<<n);i++){
            for(int j=1;j<=n;j++){
                if((i&(1<<(j-1)))==0){
                    for(int k=1;k<=n;k++){
                        if(j==k) continue;
                        if((i&(1<<(k-1)))==0){
                        //    cout<<qu[(1<<(j-1))|(1<<(k-1))]<<endl;
                            f[i|qu[(1<<(j-1))|(1<<(k-1))]]=min(f[i|qu[(1<<(j-1))|(1<<(k-1))]],f[i]+1);
                        }
                    }
                }
            }
            
            for(int j=1;j<=n;j++){
                if((i&(1<<(j-1)))==0){
                    f[i|(1<<(j-1))]=min(f[i|(1<<(j-1))],f[i]+1);
                }
            }
            
        }
//        for(int i=1;i<=(1<<n)-1;i++){
//            cout<<f[i]<<" :::"<<endl;
//        }
        cout<<f[(1<<n)-1]<<endl;
    }
    
    return 0;
}