time limit per test:2 seconds
memory limit per test:64 megabytes
input:standard input
output:standard output
We will not waste your time, it is a straightforward problem. Given multiple polygons, calculate the area of their intersection. For simplicity, there will be exactly 2 polygons both of them are convex, given in the counterclockwise order and have non-zero areas. Furthermore, in one polygon a vertex won't be on the sides of the other one. The figure below demonstrates the first test case.
Input
The first line of the input will be a single integer T, the number of test cases (1 ≤ T ≤ 20). each test case contains two integers (3 ≤ N, M ≤ 40) Then a line contains N pairs of integers xi, yi (-1000 ≤ xi, yi ≤ 1000) coordinates of the ith vertex of polygon A, followed by a line contains M pairs of integers xj, yj (-1000 ≤ xj, yj ≤ 1000) coordinates of the jth vertex of polygon B. The coordinates are separated by a single space.
Output
For each test case, print on a single line, a single number representing the area of intersection, rounded to four decimal places.
Examples
Input
2
5 3
0 3 1 1 3 1 3 5 1 5
1 3 5 3 3 6
3 3
-1 -1 -2 -1 -1 -2
1 1 2 1 1 2
Output
2.6667
0.0000
题目链接:http://codeforces.com/gym/100952/problem/J
题意:给2个凸多边形,求相交面积
思路:板子题,学习一下!
下面给出AC代码:
1 #include "iostream"
2 #include "string.h"
3 #include "stack"
4 #include "queue"
5 #include "string"
6 #include "vector"
7 #include "set"
8 #include "map"
9 #include "algorithm"
10 #include "stdio.h"
11 #include "math.h"
12 #define ll long long
13 #define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
14 #define mem(a) memset(a,0,sizeof(a))
15 #define mp(x,y) make_pair(x,y)
16 using namespace std;
17 const long long INF = 1e18+1LL;
18 const int inf = 1e9+1e8;
19 const int N=1e5+100;
20 #define maxn 510
21 const double eps=1E-8;
22 int sig(double d){
23 return(d>eps)-(d<-eps);
24 }
25 struct Point{
26 double x,y; Point(){}
27 Point(double x,double y):x(x),y(y){}
28 bool operator==(const Point&p)const{
29 return sig(x-p.x)==0&&sig(y-p.y)==0;
30 }
31 };
32 double cross(Point o,Point a,Point b){
33 return(a.x-o.x)*(b.y-o.y)-(b.x-o.x)*(a.y-o.y);
34 }
35 double area(Point* ps,int n){
36 ps[n]=ps[0];
37 double res=0;
38 for(int i=0;i<n;i++){
39 res+=ps[i].x*ps[i+1].y-ps[i].y*ps[i+1].x;
40 }
41 return res/2.0;
42 }
43 int lineCross(Point a,Point b,Point c,Point d,Point&p){
44 double s1,s2;
45 s1=cross(a,b,c);
46 s2=cross(a,b,d);
47 if(sig(s1)==0&&sig(s2)==0) return 2;
48 if(sig(s2-s1)==0) return 0;
49 p.x=(c.x*s2-d.x*s1)/(s2-s1);
50 p.y=(c.y*s2-d.y*s1)/(s2-s1);
51 return 1;
52 }
53 //多边形切割
54 //用直线ab切割多边形p,切割后的在向量(a,b)的左侧,并原地保存切割结果
55 //如果退化为一个点,也会返回去,此时n为1
56 void polygon_cut(Point*p,int&n,Point a,Point b){
57 static Point pp[maxn];
58 int m=0;p[n]=p[0];
59 for(int i=0;i<n;i++){
60 if(sig(cross(a,b,p[i]))>0) pp[m++]=p[i];
61 if(sig(cross(a,b,p[i]))!=sig(cross(a,b,p[i+1])))
62 lineCross(a,b,p[i],p[i+1],pp[m++]);
63 }
64 n=0;
65 for(int i=0;i<m;i++)
66 if(!i||!(pp[i]==pp[i-1]))
67 p[n++]=pp[i];
68 while(n>1&&p[n-1]==p[0])n--;
69 }
70 //---------------华丽的分隔线-----------------//
71 //返回三角形oab和三角形ocd的有向交面积,o是原点//
72 double intersectArea(Point a,Point b,Point c,Point d){
73 Point o(0,0);
74 int s1=sig(cross(o,a,b));
75 int s2=sig(cross(o,c,d));
76 if(s1==0||s2==0)return 0.0;//退化,面积为0
77 if(s1==-1) swap(a,b);
78 if(s2==-1) swap(c,d);
79 Point p[10]={o,a,b};
80 int n=3;
81 polygon_cut(p,n,o,c);
82 polygon_cut(p,n,c,d);
83 polygon_cut(p,n,d,o);
84 double res=fabs(area(p,n));
85 if(s1*s2==-1) res=-res;return res;
86 }
87 //求两多边形的交面积
88 double intersectArea(Point*ps1,int n1,Point*ps2,int n2){
89 if(area(ps1,n1)<0) reverse(ps1,ps1+n1);
90 if(area(ps2,n2)<0) reverse(ps2,ps2+n2);
91 ps1[n1]=ps1[0];
92 ps2[n2]=ps2[0];
93 double res=0;
94 for(int i=0;i<n1;i++){
95 for(int j=0;j<n2;j++){
96 res+=intersectArea(ps1[i],ps1[i+1],ps2[j],ps2[j+1]);
97 }
98 }
99 return res;//assumeresispositive!
100 }
101 //hdu-3060求两个任意简单多边形的并面积
102 Point ps1[maxn],ps2[maxn];
103 int n1,n2;
104 int main(){
105 int t;
106 cin>>t;
107 while(t--){
108 scanf("%d%d",&n1,&n2);
109 for(int i=0;i<n1;i++)
110 scanf("%lf%lf",&ps1[i].x,&ps1[i].y);
111 for(int i=0;i<n2;i++)
112 scanf("%lf%lf",&ps2[i].x,&ps2[i].y);
113 double ans=intersectArea(ps1,n1,ps2,n2);
114 //ans=fabs(area(ps1,n1))+fabs(area(ps2,n2))-ans;//容斥
115 printf("%.4f\n",ans);
116 }
117 return 0;
118 }