Hoj 1157 Scud busters

Question: N convex packets are given. Now we need to use a bomb to blow up the fixed coordinates. The convex hull where the attacked coordinates are located will be destroyed. Now we ask you to analyze the area of the destroyed convex hull: convex hull + Polygon Area
 

 

# Include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # include <cmath> using namespace STD; const int x = 105; # define debug puts ("here"); int top, n, a [X]; bool use [X]; struct node {int X, Y; friend bool operator <(node A, Node B) {return. Y <B. Y | (. y = B. Y &. x <B. x) ;}} res [X], p [X], in [25] [X]; int det (INT X1, int Y1, int X2, int Y2) {return X1 * y2-x2 * Y1;} bool del (INT top, int I) {If (det (RES [t OP]. x-res [Top-1]. x, Res [Top]. y-res [Top-1]. y, P [I]. x-res [Top]. x, p [I]. y-res [Top]. y) <= 0) return true; return false;} void Graham () {sort (p, p + n); Top = 1; res [0] = P [0]; Res [1] = P [1]; for (INT I = 2; I <n; I ++) {While (top & del (top, I) Top --; Res [++ top] = P [I] ;} int mint = top; res [+ TOP] = P [N-2]; for (INT I = n-3; I> = 0; I --) {While (top! = Mint & del (top, I) -- top; Res [++ top] = P [I] ;}} int det (node A, Node B, node O) {return (. x-o.x) * (B. y-o.y)-(. y-o.y) * (B. x-o.x);} bool check (node cur, int ID) {for (INT I = 0; I <A [ID]-1; I ++) if (det (in [ID] [I], in [ID] [I + 1], cur) * det (in [ID] [I + 1], in [ID] [I + 2], cur) <0) return false; return true;} double DIS (node A, Node B) {return SQRT (. x-b.x) * 1.0 * (. x-b.x) +. y-b.y) * (. y-b.y);} double CAL (int id) {double ans = 0; f Or (INT I = 1; I <A [ID]; I ++) ans + = det (in [ID] [I], in [ID] [I + 1], in [ID] [0]); ANS/= 2; ans = max (ANS,-ans); Return ans ;} double solve (int x, int y, int CNT) {node cur; cur. X = x; cur. y = y; For (INT I = 1; I <= CNT; I ++) if (! Use [I] & check (cur, I) {use [I] = true; return CAL (I) ;}return 0 ;} int main () {memset (use, false, sizeof (use); int CNT = 0; while (CIN> n, n! =-1) {for (INT I = 0; I <n; I ++) scanf ("% d", & P [I]. x, & P [I]. y); Graham (); CNT ++; For (INT I = 0; I <= top; I ++) in [CNT] [I] = res [I]; A [CNT] = top;} int X, Y; double ans = 0; while (CIN> x> Y) ans + = solve (X, Y, CNT ); printf ("%. 2lf \ n ", ANS); Return 0 ;}