POJ 2451 半平面交nlogn

题意：

半平面交面积

 

题解：

果断上模板了。

尼玛。。。我输出-0.0给wa了半天，，，抑郁。。。

 

这个好像没什么用。。。

View Code

  1 #include <iostream>

  2 #include <cstring>

  3 #include <cstdlib>

  4 #include <cstdio>

  5 #include <algorithm>

  6 #include <cmath>

  7 

  8 #define N 222222

  9 #define EPS 1e-10

 10 #define SIDE 10000

 11 

 12 using namespace std;

 13 

 14 struct PO

 15 {

 16     double x,y;

 17     void prt()

 18     {

 19         printf("%lf       %lf\n",x,y);

 20     }

 21 }p[N],o;

 22 

 23 struct LI

 24 {

 25     PO a,b;

 26     double angle;

 27     void in(double x1,double y1,double x2,double y2)

 28     {

 29         a.x=x1; a.y=y1; b.x=x2; b.y=y2;

 30     }

 31     void prt()

 32     {

 33         printf("%lf     %lf      %lf    %lf\n",a.x,a.y,b.x,b.y);

 34     }

 35 }li[N],deq[N];

 36 

 37 int n,m,cnt;

 38 

 39 inline int dc(double x)

 40 {

 41     if(x>EPS) return 1;

 42     else if(x<-EPS) return -1;

 43     return 0;

 44 }

 45 

 46 inline PO operator -(PO a,PO b)

 47 {

 48     PO c;

 49     c.x=a.x-b.x; c.y=a.y-b.y;

 50     return c;

 51 }

 52 

 53 inline double cross(PO a,PO b,PO c)

 54 {

 55     return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);

 56 }

 57 

 58 inline bool cmp(const LI &a,const LI &b)//极角排序，内侧在前 

 59 {

 60     if(dc(a.angle-b.angle)==0) return dc(cross(a.a,a.b,b.a))<0;

 61     return a.angle>b.angle;

 62 }

 63 

 64 inline PO getpoint(LI &a,LI &b)

 65 {

 66     double k1=cross(a.a,b.b,b.a);

 67     double k2=cross(a.b,b.a,b.b);

 68     PO tmp=a.b-a.a,ans;

 69     ans.x=a.a.x+tmp.x*k1/(k1+k2);

 70     ans.y=a.a.y+tmp.y*k1/(k1+k2);

 71     return ans;

 72 }

 73 

 74 inline void read()

 75 {

 76     for(int i=1;i<=n;i++)

 77     {

 78         scanf("%lf%lf%lf%lf",&li[i].a.x,&li[i].a.y,&li[i].b.x,&li[i].b.y);

 79         li[i].angle=atan2(li[i].b.y-li[i].a.y,li[i].b.x-li[i].a.x);

 80     }

 81     li[n+1].in(0,0,SIDE,0);

 82     li[n+2].in(0,SIDE,0,0);

 83     li[n+3].in(SIDE,0,SIDE,SIDE);

 84     li[n+4].in(SIDE,SIDE,0,SIDE);

 85     for(int i=n+1;i<=n+4;i++) li[i].angle=atan2(li[i].b.y-li[i].a.y,li[i].b.x-li[i].a.x);

 86     n+=4;

 87 }

 88 

 89 inline void getcut()

 90 {

 91     sort(li+1,li+1+n,cmp);

 92     m=1;

 93     for(int i=2;i<=n;i++)

 94         if(dc(li[i].angle-li[m].angle)!=0)

 95             li[++m]=li[i];

 96     deq[1]=li[1]; deq[2]=li[2];

 97     int bot=1,top=2;

 98     for(int i=3;i<=m;i++)

 99     {

100         while(bot<top&&dc(cross(li[i].a,li[i].b,getpoint(deq[top],deq[top-1])))<0) top--;

101         while(bot<top&&dc(cross(li[i].a,li[i].b,getpoint(deq[bot],deq[bot+1])))<0) bot++;

102         deq[++top]=li[i];

103     }

104     while(bot<top&&dc(cross(deq[bot].a,deq[bot].b,getpoint(deq[top],deq[top-1])))<0) top--;

105     while(bot<top&&dc(cross(deq[top].a,deq[top].b,getpoint(deq[bot],deq[bot+1])))<0) bot++;

106     if(bot==top) return;

107     cnt=0;

108     for(int i=bot;i<top;i++) p[++cnt]=getpoint(deq[i],deq[i+1]);

109     if(top-1>bot) p[++cnt]=getpoint(deq[bot],deq[top]);

110 }

111 

112 inline double getarea()

113 {

114     double res=0.0;

115     p[cnt+1]=p[1];

116     for(int i=1;i<=cnt;i++) res+=cross(o,p[i],p[i+1]);

117     return res;

118 }

119 

120 inline void go()

121 {

122     getcut();

123     printf("%.1lf\n",fabs(getarea())*0.5);

124 }

125 

126 int main()

127 {

128     while(scanf("%d",&n)!=EOF) read(),go();

129     return 0;

130 }