• bzoj 2732: [HNOI2012]射箭


    额,,一看就设抛物线是y=a*x^2+b*x,把x,y1,y2带进去就是关于a,b的不等式。。然后半平面交。。。

    考虑答案,可以二分。

    (然而还是不对,,被卡精度了23333,(话说百度卡常数,,直接连膝盖都没了。。),,而且开始的时候异常自信,直接没加边界,就搞搞搞23333,闷声做大死2333)

     1 #include<bits/stdc++.h>
     2 #define N 100005
     3 #define LL long long 
     4 #define eps 1e-8
     5 #define double long double
     6 using namespace std;
     7 inline int ra()
     8 {
     9     int x=0,f=1; char ch=getchar();
    10     while (ch<'0' || ch>'9') {if (ch=='-') f=-1; ch=getchar();}
    11     while (ch>='0' && ch<='9') {x=x*10+ch-'0'; ch=getchar();}
    12     return x*f;
    13 }
    14 int n,cnt,tot;
    15 const double linf=1e15;
    16 struct in_data{double x,y1,y2;} q[N];
    17 struct point{double x,y;};
    18 struct line{
    19     point a,b; double angle;
    20     void print()
    21     {
    22         printf("%.1lf   %.1lf   %.1lf   %.1lf
    ",a.x,a.y,b.x,b.y);
    23     }
    24 } l[N<<1],s[N<<1];
    25 double operator * (point a, point b){
    26     return a.x*b.y-a.y*b.x;
    27 }
    28 point operator - (point a, point b){
    29     point t; t.x=a.x-b.x; t.y=a.y-b.y; return t;
    30 }
    31 bool operator < (line a, line b){
    32     if (a.angle==b.angle) return (b.b-a.a)*(a.b-a.a)>0;
    33     return a.angle<b.angle;
    34 }
    35 point intersection(line a, line b)
    36 {
    37     double k1,k2,t; point ans;
    38     k1=(b.a-a.a)*(a.b-a.a);
    39     k2=(a.b-a.a)*(b.b-a.a);
    40     t=k1/(k1+k2);
    41     ans.x=b.a.x+(b.b.x-b.a.x)*t;
    42     ans.y=b.a.y+(b.b.y-b.a.y)*t;
    43     return ans;
    44 }
    45 bool jud(line a, line b, line t){
    46     point p=intersection(a,b);
    47     return (t.a-p)*(t.b-p)<0;
    48 }
    49 bool half_plane_intersection()
    50 {
    51     tot=0;
    52     for (int i=1; i<=cnt; i++)
    53         if (l[i].angle!=l[i-1].angle) l[++tot]=l[i];
    54     int start=1,end=0; cnt=tot;
    55     s[++end]=l[1]; s[++end]=l[2];
    56     for (int i=3; i<=cnt; i++)
    57     {
    58         while (start<end && jud(s[end],s[end-1],l[i])) end--;
    59         while (start<end && jud(s[start],s[start+1],l[i])) start++;
    60         s[++end]=l[i];
    61     } 
    62     while (start<end && jud(s[end],s[end-1],s[start])) end--;
    63     while (start<end && jud(s[start],s[start+1],s[end])) start++;
    64     return end-start>=2;
    65 }
    66 bool check(int len)
    67 {
    68     cnt=0;
    69     if (len<=2) return 1;
    70     for (int i=1; i<=len; i++)
    71     {
    72         l[++cnt].a.x=-1; l[cnt].a.y=q[i].y1/q[i].x+q[i].x;
    73         l[cnt].b.x=1; l[cnt].b.y=q[i].y1/q[i].x-q[i].x;
    74         l[++cnt].a.x=1; l[cnt].a.y=q[i].y2/q[i].x-q[i].x;
    75         l[cnt].b.x=-1; l[cnt].b.y=q[i].y2/q[i].x+q[i].x;
    76     }
    77     l[++cnt].a=(point){-linf,-linf};l[cnt].b=(point){linf,-linf};
    78     l[++cnt].a=(point){linf,-linf};l[cnt].b=(point){linf,linf};
    79     l[++cnt].a=(point){linf,linf};l[cnt].b=(point){-linf,linf};
    80     l[++cnt].a=(point){-linf,linf};l[cnt].b=(point){-linf,-linf};
    81     for (int i=1; i<=cnt; i++)
    82         l[i].angle=atan2(l[i].b.y-l[i].a.y,l[i].b.x-l[i].a.x);
    83     sort(l+1,l+cnt+1);
    84     return half_plane_intersection();
    85 }
    86 int main()
    87 {
    88     n=ra();
    89     for (int i=1; i<=n; i++) q[i].x=ra(),q[i].y1=ra(),q[i].y2=ra();
    90     int L=1,R=n,ans;
    91     while (L<=R)
    92     {
    93         int mid=L+R>>1;
    94         if (check(mid)) ans=mid,L=mid+1; 
    95             else R=mid-1; 
    96     }
    97     cout<<ans;
    98     return 0;
    99 }
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  • 原文地址:https://www.cnblogs.com/ccd2333/p/6492689.html
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