hdu2297Run（凸包的巧妙应用）

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很巧妙的一道题，参考

把距离和速度分别作为x和y坐标，以斜率代表追赶速率，简直炫酷~

具体看上面的博客，画的很清楚，就不再抄写一遍了。

  1 #include <iostream>

  2 #include<cstdio>

  3 #include<cstring>

  4 #include<algorithm>

  5 #include<stdlib.h>

  6 #include<vector>

  7 #include<cmath>

  8 #include<queue>

  9 #include<set>

 10 using namespace std;

 11 #define N 50010

 12 #define LL long long

 13 #define INF 0xfffffff

 14 const double eps = 1e-8;

 15 const double pi = acos(-1.0);

 16 const double inf = ~0u>>2;

 17 struct point

 18 {

 19     double x,y;

 20     point(double x=0,double y =0 ):x(x),y(y){}

 21 }p[N],ch[N];

 22 typedef point pointt;

 23 point operator -(point a,point b)

 24 {

 25     return point(a.x-b.x,a.y-b.y);

 26 }

 27 double cross(point a,point b)

 28 {

 29     return a.x*b.y-a.y*b.x;

 30 }

 31 int dcmp(double x)

 32 {

 33     if(fabs(x)<eps) return 0;

 34      else return x<0?-1:1;

 35 }

 36 double mul(point p0,point p1,point p2)

 37 {

 38     return cross(p1-p0,p2-p0);

 39 }

 40 double dis(point a)

 41 {

 42     return sqrt(a.x*1.0*a.x+a.y*1.0*a.y);

 43 }

 44 bool cmp(point a,point b)

 45 {

 46     if(dcmp(mul(p[0],a,b))==0)

 47         return dis(a-p[0])<dis(b-p[0]);

 48     else

 49         return dcmp(mul(p[0],a,b))>0;

 50 }

 51 int Graham(int n)

 52 {

 53     if(n<2) return n;

 54     int i,k = 0,top;

 55     point tmp;

 56     for(i = 0 ; i < n; i++)

 57     {

 58         if(p[i].y<p[k].y||(p[i].y==p[k].y&&p[i].x<p[k].x))

 59             k = i;

 60     }

 61     if(k!=0)

 62     {

 63         tmp = p[0];

 64         p[0] = p[k];

 65         p[k] = tmp;

 66     }

 67     sort(p+1,p+n,cmp);

 68     ch[0] = p[0];

 69     ch[1] = p[1];

 70     top = 1;

 71     for(i = 2; i < n ; i++)

 72     {

 73         while(top>0&&dcmp(mul(ch[top-1],ch[top],p[i]))<=0)

 74             top--;

 75         top++;

 76         ch[top] = p[i];

 77     }

 78     return top;

 79 }

 80 int main()

 81 {

 82     int t,i,n;

 83     cin>>t;

 84     while(t--)

 85     {

 86         scanf("%d",&n);

 87         for(i = 0 ; i < n; i++)

 88         scanf("%lf%lf",&p[i].y,&p[i].x);

 89         int m = Graham(n);

 90         int maxx = -INF,tx,maxy = -INF,ty;

 91         if(n<2)

 92         {

 93             cout<<"1\n";

 94             continue;

 95         }

 96         for(i = 0 ; i <= m ; i++)

 97         {

 98             //cout<<ch[i].x<<" "<<ch[i].y<<endl;

 99             if(ch[i].x>maxx)

100             {

101                 tx = i;

102                 maxx = ch[i].x;

103             }

104             if(ch[i].y>maxy)

105             {

106                 ty = i;

107                 maxy = ch[i].y;

108             }

109         }

110         cout<<ty-tx+1<<endl;

111     }

112     return 0;

113 }

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