本文主要是介绍FZOJ2110 star(DFS)，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
Overpower often go to the playground with classmates. They play and chat on the playground. One day, there are a lot of stars in the sky. Suddenly, one of Overpower’s classmates ask him: “How many acute triangles whose inner angles are less than 90 degrees (regarding stars as points) can be found? Assuming all the stars are in the same plane”. Please help him to solve this problem.
Input
The first line of the input contains an integer T (T≤10), indicating the number of test cases.
For each test case:
The first line contains one integer n (1≤n≤100), the number of stars.
The next n lines each contains two integers x and y (0≤|x|, |y|≤1,000,000) indicate the points, all the points are distinct.
Output
For each test case, output an integer indicating the total number of different acute triangles.
Sample Input
1 3 0 0 10 0 5 1000
Sample Output
1
#include<stdio.h> #include<math.h> typedef struct nn {double x1,y1; }node; void cmp(double *a,double *b,double *c) {double tem;if(*a<*b){tem=*a;*a=*b;*b=tem;}if(*a<*c){tem=*a;*a=*c;*c=tem;} } double cacreat(double x1,double y1,double x2,double y2) {double edglen;edglen=sqrt(pow(x1-x2,2.0)+pow(y1-y2,2.0));return edglen; } int pandu(double a,double b,double c) {if(b*b+c*c-a*a>0)return 1;return 0; } double x[105],y[105]; node s[3]; int vist[105],n,sum; void dfs(int cout,int j) {double edg1,edg2,edg3;int i;s[cout].x1=x[j];s[cout].y1=y[j];if(cout==2){edg1=cacreat(s[0].x1,s[0].y1,s[1].x1,s[1].y1);edg2=cacreat(s[0].x1,s[0].y1,s[2].x1,s[2].y1);edg3=cacreat(s[1].x1,s[1].y1,s[2].x1,s[2].y1);cmp(&edg1,&edg2,&edg3);if(pandu(edg1,edg2,edg3)!=0)sum++;return ;}for(i=j+1;i<n;i++)if(vist[i]==0){vist[i]=1;dfs(cout+1,i);vist[i]=0;} } int main() {int i,t;scanf("%d",&t);while(t--){scanf("%d",&n);for(i=0;i<n;i++){scanf("%lf%lf",&x[i],&y[i]);vist[i]=0;}sum=0;for(i=0;i<n;i++){vist[i]=1;dfs(0,i);vist[i]=0;}printf("%d\n",sum);} }
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