Fzu problem 2110 star

Enumeration is fine, because we have just learned DFS, so what we do with DFS is actually three for loops.

 

#include<cstdio>#include<cmath>#include<cstring>#define HG  -1e-12using namespace std;int n,sum,vis[110],temp[5];typedef struct point{    int x,y;} point;point p[110];int isok(int x1,int y1,int x2,int y2,int x3,int y3 ){    double a=sqrt((y2-y1)*1.0*(y2-y1)+(x2-x1)*1.0*(x2-x1));    double b=sqrt((y3-y1)*1.0*(y3-y1)+(x3-x1)*1.0*(x3-x1));    double c=sqrt((y3-y2)*1.0*(y3-y2)+(x3-x2)*1.0*(x3-x2));    double C=(a*a+b*b-c*c)/(2*a*b);    double B=(a*a+c*c-b*b)/(2*a*c);    double A=(b*b+c*c-a*a)/(2*b*c);    if(A>HG&&B>HG&&C>HG)        return 1;    else return 0;}void dfs(int cur,int num){    vis[cur]=1;    temp[num]=cur;    if(num==2)    {        if(isok(p[temp[0]].x,p[temp[0]].y,p[temp[1]].x,p[temp[1]].y,p[temp[2]].x,p[temp[2]].y))            sum++;    }    else for(int i=0; i<n; i++)        {            if(!vis[i]&&i>cur)            {                dfs(i,num+1);                vis[i]=0;            }        }}int main(){    //freopen("in.txt","r",stdin);    int cas;    scanf("%d",&cas);    while(cas--)    {        scanf("%d",&n);        sum=0;        for(int i=0; i<n; i++)            scanf("%d%d",&p[i].x,&p[i].y);        memset(vis,0,sizeof(vis));        for(int i=0; i<n; i++)        {            memset(vis,0,sizeof(vis));            dfs(i,0);        }        printf("%d\n",sum);    }    return 0;}

The first Wa is only conducted DFS (0, 0), no DFS starting from the second point, the second is to determine whether it is an acute triangle, double A = SQRT (y2-y1) * 1.0 * (y2-y1) + (x2-x1) * 1.0 * (x2-x1); this is because the x, y range is 1000000 and will overflow if it is saved with int, so * is converted to double.

"Higher Education Community Cup" third Fujian University Student Program Design Competition