题意：给定一些半平面，求面积。。
直接套用模板。。
code：
1 /* 2 Time:2013-04-11 19:09:39 3 State:Accepted 4 */ 5 #include<iostream> 6 #include<cstring> 7 #include<cstdio> 8 #include<cmath> 9 #include<algorithm> 10 #include<set> 11 #define eps 1e-8 12 #define maxn 25010 13 using namespace std; 14 15 int ln, q[maxn], top, bot, n, ord[maxn]; 16 17 struct point{double x, y; } p[maxn]; 18 struct line { 19 point a, b; 20 double angle; 21 } l[maxn]; 22 23 void add_line(double x1, double y1, double x2, double y2 ){ 24 l[ln].a.x = x1; 25 l[ln].b.x = x2; 26 l[ln].a.y = y1; 27 l[ln].b.y = y2; 28 l[ln].angle = atan2(y2 - y1, x2 - x1); 29 ord[ln] = ln; 30 ++ln; 31 } 32 33 void init(){ 34 double x1 , y1, x2, y2; 35 while (scanf("%d", &n) != EOF){ 36 for (int i = ln = 0; i < n; ++i){ 37 scanf("%lf%lf%lf%lf",&x1, &y1, &x2, &y2); 38 add_line(x1 , y1, x2, y2); 39 } 40 } 41 add_line(0,0,10000,0); 42 add_line(10000,0,10000,10000); 43 add_line(10000,10000,0,10000); 44 add_line(0,10000,0,0); 45 } 46 47 int dblcmp(double k){ 48 if (fabs(k) < eps) return 0; 49 return k > 0 ? 1 : -1; 50 } 51 52 double multi(point p0, point p1, point p2){ 53 return (p1.x-p0.x)*(p2.y-p0.y)-(p1.y-p0.y)*(p2.x-p0.x); 54 } 55 56 bool cmp(const int u, const int v){ 57 int d = dblcmp(l[u].angle - l[v].angle); 58 if (d == 0) return dblcmp(multi(l[u].a, l[v].a,l[v].b)) > 0; 59 return d < 0; 60 } 61 62 void get_point(line l1, line l2, point& p){ 63 double dot1, dot2; 64 dot1 = multi(l2.a, l1.b, l1.a); 65 dot2 = multi(l1.b, l2.b, l1.a); 66 p.x = (l2.a.x * dot2 + l2.b.x * dot1) / (dot1 + dot2); 67 p.y = (l2.a.y * dot2 + l2.b.y * dot1) / (dot1 + dot2); 68 } 69 70 bool judge(line l0, line l1, line l2){ 71 point p; 72 get_point(l1, l2, p); 73 return dblcmp(multi(p, l0.a , l0.b)) < 0; 74 } 75 76 void SAI(){ 77 sort(ord, ord + ln , cmp); 78 int i, j; 79 for (i = 0, j = 0; i < ln; ++i) 80 if (dblcmp(l[ord[i]].angle - l[ord[j]].angle) > 0) 81 ord[++j] = ord[i]; 82 ln = j + 1; 83 q[0] = ord[0]; 84 q[1] = ord[1]; 85 /* for (int i = 0; i < ln; ++i) 86 printf("order[%d] = %d\n", i, ord[i]);*/ 87 88 bot = 0; 89 top = 1; 90 for (int i = 2; i < ln ; ++i){ 91 while (bot < top && judge(l[ord[i]], l[q[top - 1]], l[q[top]])) --top; 92 while (bot < top && judge(l[ord[i]], l[q[bot + 1]], l[q[bot]])) ++bot; 93 q[++top] = ord[i]; 94 } 95 while (bot < top && judge(l[q[bot]], l[q[top-1]], l[q[top]])) --top; 96 while (bot < top && judge(l[q[top]], l[q[bot + 1]], l[q[bot]])) ++bot; 97 q[++top] = q[bot]; 98 // printf("top = %d bot = %d\n", top, bot); 99 /* for (int i = bot; i < top; ++i) 100 printf("q[%d] = %d\n", i, q[i]);*/ 101 102 for (n = 0, i = bot; i < top; i++, n++) 103 get_point(l[q[i+1]], l[q[i]], p[n]); 104 } 105 106 double getArea() { 107 if (n < 3) return 0; 108 double area = 0; 109 for (int i = 1; i < n-1; i++) 110 area += multi(p[0], p[i], p[i+1]); 111 return fabs(area)/2; 112 } 113 114 int main(){ 115 freopen("poj2451.in","r",stdin); 116 freopen("poj2451.out","w",stdout); 117 init(); 118 SAI(); 119 printf("%.1lf\n",getArea()); 120 fclose(stdin); fclose(stdout); 121 }