NOIP2016提高组day2 愤怒的小鸟

Description

Data Constraint

Solution

还是一道dp题。这次是一道状态压缩dp。我们先预处理出任意选择两个点构成的抛物线所覆盖的点，然后做一下状态压缩转移一下就好了。要注意卡一下精度。

Code

#include
#include
#include
#include
#include
#define db double
using namespace std;
const int maxn=20,maxn1=524288+5;
const db mo=1e-10;
db x[maxn],y[maxn],a,b,z;
int n,m,i,t,j,k,l,p,q,f[maxn1],c[maxn][maxn],er[maxn];
int main(){
    freopen("angrybirds.in","r",stdin);freopen("angrybirds.out","w",stdout);
    scanf("%d",&l);er[1]=1;
    for (i=2;i<=maxn;i++)
        er[i]=er[i-1]*2;
    while (l){
        l--;
        scanf("%d%d",&n,&m);
        for (i=1;i<=n;i++)
            scanf("%lf%lf",&x[i],&y[i]);
        memset(c,0,sizeof(c));
        for (i=1;i<=n;i++)
            for (j=i+1;j<=n;j++){
                z=y[j]-x[j]*y[i]/x[i];
                a=z/x[j]/(x[j]-x[i]);
                b=(y[i]-x[i]*x[i]*a)/x[i];
                if (a>=0) continue;
                for (k=1;k<=n;k++)
                    if (abs(a*x[k]*x[k]+b*x[k]-y[k])<=mo) c[i][j]+=er[k];
            }
        memset(f,127,sizeof(f));f[0]=0;p=(1<1;q=f[1];
        for (i=0;iif (f[i]==q) continue;
            for (j=1;j<=n;j++)
                if (!(er[j] & i)){
                    f[i+er[j]]=min(f[i+er[j]],f[i]+1);
                    for (k=j+1;k<=n;k++)
                        if (!(er[k] & i)){
                            t=(i | c[j][k]);
                            f[t]=min(f[t],f[i]+1);
                        }
                }

        }
        printf("%d\n",f[p]);
    }
    return 0;
}