Moon Game （凸四边形个数，数学题）
Problem 2148 Moon Game
Accept: 24 Submit: 61 Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem Description
Fat brother and Maze are playing a kind of special (hentai) game in the clearly blue sky which we can just consider as a kind of two-dimensional plane. Then Fat brother starts to draw N starts in the sky which we can just consider each as a point. After he draws these stars, he starts to sing the famous song “The Moon Represents My Heart” to Maze.
You ask me how deeply I love you,
How much I love you?
My heart is true,
My love is true,
The moon represents my heart.
…
But as Fat brother is a little bit stay-adorable(呆萌), he just consider that the moon is a special kind of convex quadrilateral and starts to count the number of different convex quadrilateral in the sky. As this number is quiet large, he asks for your help.
Input
The first line of the date is an integer T, which is the number of the text cases.
Then T cases follow, each case contains an integer N describe the number of the points.
Then N lines follow, Each line contains two integers describe the coordinate of the point, you can assume that no two points lie in a same coordinate and no three points lie in a same line. The coordinate of the point is in the range[-10086,10086].
1 <= T <=100, 1 <= N <= 30
Output
For each case, output the case number first, and then output the number of different convex quadrilateral in the sky. Two convex quadrilaterals are considered different if they lie in the different position in the sky.
Sample Input
Sample Output
#include<stdio.h> #include<math.h> #include<string.h> #define MAX 35 __int64 ans[MAX][2]; int cheak(int i,int j,int k,int l) { __int64 x1,y1,x2,y2,x3,y3,x4,y4; x1=ans[i][0]; y1=ans[i][1]; x2=ans[j][0]; y2=ans[j][1]; x3=ans[k][0]; y3=ans[k][1]; x4=ans[l][0]; y4=ans[l][1]; double k1,k2,b1,b2,s1,s2,s3,s4; if(x2==x1&&x3==x4) return 0; if(x2==x1&&x3!=x4) { s1=x3-x1; s2=x4-x1; if(s1*s2>=0) return 0; else { k2=(y3-y4)*1.0/(x3-x4); b2=y3-k2*x3; s3=(y2-b2)-k2*x2; s4=(y1-b2)-k2*x1; if(s3*s4>=0) return 0; if(s1*s2<0&&s3*s4<0) return 1; } } if(x3==x4&&x1!=x2) { s3=x1-x3; s4=x2-x3; if(s3*s4>=0) return 0; else { k1=(y2-y1)*1.0/(x2-x1); b1=y1-k1*x1; s1=(y3-b1)-k1*x3; s2=(y4-b1)-k1*x4; if(s1*s2>=0) return 0; if(s1*s2<0&&s3*s4<0) return 1; } } else { k1=(y2-y1)*1.0/(x2-x1); b1=y1-k1*x1; s1=(y3-b1)-k1*x3; s2=(y4-b1)-k1*x4; if(s1*s2>=0) { return 0; } else { k2=(y3-y4)*1.0/(x3-x4); b2=y3-k2*x3; s3=(y2-b2)-k2*x2; s4=(y1-b2)-k2*x1; if(s3*s4>=0) return 0; if(s1*s2<0&&s3*s4<0) return 1; } } } int main() { int T,N; int i,j,k,l,cnt; scanf("%d",&T); for(cnt=1;cnt<=T;cnt++) { scanf("%d",&N); memset(ans,0,sizeof(ans)); for(i=0;i<N;i++) { scanf("%I64d%I64d",&ans[i][0],&ans[i][1]); } int temp=0; for(i=0;i<N-3;i++) for(j=i+1;j<N-2;j++) for(k=j+1;k<N-1;k++) for(l=k+1;l<N;l++) { if(cheak(i,j,k,l)) temp++; if(cheak(i,k,j,l)) temp++; if(cheak(i,l,k,j)) temp++; } printf("Case %d: %d\n",cnt,temp); } return 0; } /* 20 4 0 0 100 0 0 100 100 100 5 0 0 10 0 0 5 5 5 15 0 4 0 0 100 0 0 100 10 10 5 1 1 5 6 9 7 5 5 2 2 4 1 1 1 1 1 1 1 1 5 1 1 2 2 3 3 4 4 5 5 6 5 8 6 4 1 0 10 5 8 6 3 5 */
表示比赛的时候策略错误。。。应该让我自己做这题的，不过还是很感谢我给力的队友啊。。。
（虽然我是个小坑！ o(∩_∩)o 哈哈）那么我会好好努力的，大家一起加油哈！
