[NOI2010]：成长快乐

传送门
目前30 at 12.14
话说洛谷此题样例错了，而且第二个点是第十个点。。。
首先第一个点可以手算，10分
第二个点爆搜：

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>#include<cmath>#define ll long long#define min(a,b) a<b?a:b#define max(a,b) a>b?a:busing namespace std;const int N=1e5+5;const double eps=1e-12;struct Point{    double x,y;    Point(){}    Point(double _x,double _y):x(_x),y(_y){}    inline Point operator + (const Point& b) const {return Point(x+b.x,y+b.y);}    inline Point operator - (const Point& b) const {return Point(x-b.x,y-b.y);}    inline Point operator * (double b) const {return Point(x*b,y*b);}}nemo,P[N],v[N];inline int dcmp(double x){    if(fabs(x)<eps)return 0;    return x>eps?1:-1;}inline double dis(const Point& a,const Point& b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}inline double dis2(const Point& a,const Point& b){return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}inline double Dot(const Point& a,const Point& b){return a.x*b.x+a.y*b.y;}inline double Mod(const Point& a){return sqrt(Dot(a,a));}inline double Mod2(const Point& a){return Dot(a,a);}int n,s[N],use[N];double w0,V,T,w[N];double maxw;int ans[N],anscnt;double tm[N],anstm[N];Point pos[N],anspos[N];inline double solve(double a,double b,double c){    double d=b*b-4.0*a*c;    if(dcmp(d)<0)return -1;    d+=eps;d=sqrt(d);    double x1=(-b+d)/(2.0*a),x2=(-b-d)/(2.0*a);    double x=max(x1,x2);    if(x<0)return -1;    if(x1<0)x1=1e9;    if(x2<0)x2=1e9;    x=min(x1,x2);    return x;}inline void dfs(int cnt,double noww,double nowt,Point now){    if(nowt>T)return;    if(noww>maxw){        for(int i=1;i<cnt;i++)ans[i]=s[i];        for(int i=1;i<cnt;i++)anstm[i]=tm[i];        for(int i=1;i<cnt;i++)anspos[i]=pos[i];        maxw=noww;anscnt=cnt;    }    if(cnt>n)return;    for(int i=1;i<=n;i++){        if(use[i])continue;        if(w[i]>=noww+eps)continue;        Point B=P[i]+v[i]*nowt,A=now;        double a=Mod2(v[i])-V*V;        double b=-2.0*Dot(v[i],A-B);        double c=dis2(A,B);        double tmp=solve(a,b,c);        if(tmp==-1)continue;        use[i]=1;s[cnt]=i;tm[cnt]=tmp+nowt;pos[cnt]=B+v[i]*tmp;        dfs(cnt+1,noww+w[i],nowt+tmp,pos[cnt]);        use[i]=0;    }}int main(){    freopen("nemo2.in","r",stdin);    freopen("nemo2.out","w",stdout);    scanf("%lf %lf %lf %lf %lf",&w0,&V,&T,&nemo.x,&nemo.y);scanf("%d",&n);    for(int i=1;i<=n;i++)scanf("%lf %lf %lf %lf %lf",&w[i],&P[i].x,&P[i].y,&v[i].x,&v[i].y);    dfs(1,w0,0,nemo);    printf("%d\n",anscnt-1);    printf("%.6lf\n",maxw-w0);    for(int i=1;i<anscnt;i++)printf("%.6lf %.6lf %.6lf %d\n",anstm[i],anspos[i].x,anspos[i].y,ans[i]);    return 0;}

第三个点因为每次只能吃一个，所以排序然后模拟即可：

#include<cstdio>#include<cstdlib>#include<algorithm>#include<cstring>#include<cmath>#define ll long long#define min(a,b) a<b?a:b#define max(a,b) a>b?a:busing namespace std;const int N=1e5+5;const double eps=1e-12;struct Point{    double x,y;    Point(){}    Point(double _x,double _y):x(_x),y(_y){}    inline Point operator + (const Point& b) const {return Point(x+b.x,y+b.y);}    inline Point operator - (const Point& b) const {return Point(x-b.x,y-b.y);}    inline Point operator * (double b) const {return Point(x*b,y*b);}}nemo,P[N],v[N];struct fish{    Point pos,v;double w;int id;}F[N];inline bool cmp(const fish& a,const fish& b){return a.w<b.w;}inline int dcmp(double x){    if(fabs(x)<eps)return 0;    return x>eps?1:-1;}inline double dis(const Point& a,const Point& b){return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}inline double dis2(const Point& a,const Point& b){return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);}inline double Dot(const Point& a,const Point& b){return a.x*b.x+a.y*b.y;}inline double Mod(const Point& a){return sqrt(Dot(a,a));}inline double Mod2(const Point& a){return Dot(a,a);}int n,s[N],use[N];double w0,V,T,w[N];double tm[N];Point pos[N];inline double solve(double a,double b,double c){    double d=b*b-4.0*a*c;    if(dcmp(d)<0)return -1;    d+=eps;d=sqrt(d);    double x1=(-b+d)/(2.0*a),x2=(-b-d)/(2.0*a);    double x=max(x1,x2);    if(x<0)return -1;    if(x1<0)x1=1e9;    if(x2<0)x2=1e9;    x=min(x1,x2);    return x;}int main(){    freopen("nemo4.in","r",stdin);    freopen("nemo4.out","w",stdout);    scanf("%lf %lf %lf %lf %lf",&w0,&V,&T,&nemo.x,&nemo.y);scanf("%d",&n);    for(int i=1;i<=n;i++)scanf("%lf %lf %lf %lf %lf",&F[i].w,&P[i].x,&P[i].y,&v[i].x,&v[i].y)    ,F[i].pos=P[i],F[i].v=v[i],F[i].id=i;    sort(F+1,F+n+1,cmp);    int cnt=1;double noww=w0,nowt=0;Point now=nemo;    for(int i=1;i<=n;i++){        Point B=F[i].pos+F[i].v*nowt,A=now;        double a=Mod2(F[i].v)-V*V;        double b=-2.0*Dot(F[i].v,A-B);        double c=dis2(A,B);        double tmp=solve(a,b,c);        s[cnt]=F[i].id;tm[cnt]=tmp+nowt;pos[cnt]=B+F[i].v*tmp;        noww+=F[i].w;nowt+=tmp;now=pos[cnt];        cnt++;    }    printf("%d\n",cnt-1);    printf("%.6lf\n",noww-w0);    for(int i=1;i<cnt;i++)printf("%.6lf %.6lf %.6lf %d %.6lf\n",tm[i],pos[i].x,pos[i].y,s[i],F[i].w);    return 0;}