lightoj1418Trees on My Island【pick公式】

Q - Trees on My Island

Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1418

Description

I have bought an island where I want to plant trees in rows and columns. So, the trees will form a rectangular grid and each of them can be thought of having integer coordinates by taking a suitable grid point as the origin.

But, the problem is that the island itself is not rectangular. So, I have identified a simple polygonal area inside the island with vertices on the grid points and have decided to plant trees on grid points lying strictly inside the polygon.

Figure: A sample of my island

For example, in the above figure, the green circles form the polygon, and the blue circles show the position of the trees.

Now, I seek your help for calculating the number of trees that can be planted on my island.

Input

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing an integer N (3 ≤ N ≤ 10000) denoting the number of vertices of the polygon. Each of the next N lines contains two integers x_i y_i(-10⁶ ≤ x_i, y_i ≤ 10⁶) denoting the co-ordinate of a vertex. The vertices will be given in clockwise or anti-clockwise order. And they will form a simple polygon.

Output

For each case, print the case number and the total number of trees that can be planted inside the polygon.

Sample Input

1

9

1 2

2 1

4 1

4 3

6 2

6 4

4 5

1 5

2 3

Sample Output

Case 1: 8

题意：求多边形内的点数

解题思路：pick公式

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>using namespace std;struct point{long long x,y;}A[10010];long long gcd(long long a,long long b){return b==0?a:gcd(b,a%b);}int main(){long long t,k=1,n,i,j;scanf("%lld",&t);while(t--){scanf("%lld",&n);for(i=0;i<n;++i){scanf("%lld%lld",&A[i].x,&A[i].y);}long long dot=0,S=0;for(i=0;i<n;++i){dot+=gcd(abs(A[(i+1)%n].x-A[i].x),abs(A[(i+1)%n].y-A[i].y));S+=A[(i+1)%n].y*A[i].x-A[i].y*A[(i+1)%n].x;}printf("Case %lld: %lld\n",k++,(abs(S)+2-dot)/2);}return 0;}