FZU - 2148 Moon Game(几何)
2017-08-10 17:12
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原题:
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
译
题意:
给出n个点坐标,求一共可以组成多少个凸四边形。
思路:
由于测试量很小,完全可以所有点暴力枚举,在判断是否是凸四边形时,可以利用叉积计算四个顶点中每三个顶点组成的三角形面积,如果不是凸四边形,那么最大的三角形面积一定是另外三个三角形面积之和。
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
2 4 0 0 100 0 0 100 100 100 4 0 0 100 0 0 100 10 10
Sample Output
Case 1: 1 Case 2: 0
译
题意:
给出n个点坐标,求一共可以组成多少个凸四边形。
思路:
由于测试量很小,完全可以所有点暴力枚举,在判断是否是凸四边形时,可以利用叉积计算四个顶点中每三个顶点组成的三角形面积,如果不是凸四边形,那么最大的三角形面积一定是另外三个三角形面积之和。
#include <iostream> #include <iomanip> #include <algorithm> #include <cstdio> #include <cstring> #include <queue> #include <deque> #include <string> #include <cmath> #include <vector> #include <utility> #include <set> #include <map> #include <climits> //#pragma comment(linker, "/STACK:1024000000,1024000000") #define INF 2147483647 using namespace std; typedef long long ll; int T,i,n,j,t,k,l; struct PPoint { int x,y; } pp[35]; double Aarea(int a,int b,int c) { double s=(double)(1.0/2)*(pp[a].x*pp[b].y+pp[b].x*pp[c].y+pp[c].x*pp[a].y-pp[a].x*pp[c].y-pp[b].x*pp[a].y-pp[c].x*pp[b].y); if(s<0) return -s; return s; } int main() { scanf("%d",&T); for(t=1; t<=T; t++) { memset(pp,0,sizeof(pp)); scanf("%d",&n); for(i=0; i<n; i++) scanf("%d%d",&pp[i].x,&pp[i].y); int res=0; for(i=0; i<n; i++) for(j=i+1; j<n; j++) for(k=j+1; k<n; k++) for(l=k+1; l<n; l++) { double are[10]; memset(are,0,sizeof(are)); are[0]=Aarea(i,j,k); are[1]=Aarea(i,j,l); are[2]=Aarea(j,k,l); are[3]=Aarea(k,l,i); sort(are,are+4); if(are[3]-are[1]-are[2]-are[0]<0.01&&are[3]-are[1]-are[2]-are[0]>-0.01) continue; res++; } printf("Case %d: %d\n",t,res); } return 0; }
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