poj1244Slots of Fun

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几何的简单题，建立坐标，判断相等以及不共线

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 200
#define LL long long
#define INF 0xfffffff
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
char s[N],sr[30];
struct Point
{
    double x,y;
    Point(double x=0,double y=0):x(x),y(y) {}
}p[N];
vector<Point>ed[30];
typedef Point pointt;
pointt operator + (Point a,Point b)
{
    return Point(a.x+b.x,a.y+b.y);
}
pointt operator - (Point a,Point b)
{
    return Point(a.x-b.x,a.y-b.y);
}
int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    else return x<0?-1:1;
}
double dis(Point a)
{
    return sqrt(a.x*a.x+a.y*a.y);
}
double cross(Point a,Point b)
{
    return a.x*b.y-a.y*b.x;
}
double mul(Point p0,Point p1,Point p2)
{
    return cross(p1-p0,p2-p0);
}
int main()
{
    int n,i,j,g,e;
    while(scanf("%d",&n)&&n)
    {
        getchar();
        for(i = 0; i < 26 ; i++)
        ed[i].clear();
        for(i = 1 ; i <= n*(n+1)/2; i++)
        scanf("%c",&s[i]);
        int g = 0;
        double tx=n,ty = sqrt(3.0)*(n-1);
        for(i = 1 ;i <= n; i++)
        {
            double x0 = tx-i+1;
            double y0 = ty-(i-1)*sqrt(3.0);
            for(j = 1; j <= i ;j++)
            {
                Point pp;
                pp.y = y0;
                pp.x = x0;
                x0+=2;
                g++;
                ed[s[g]-'a'].push_back(pp);
            }
        }
        int cnt = 0;
        for(i = 0; i < 26 ; i++)
        {
            if(ed[i].size()<3) continue;
            int k = ed[i].size();
            for(j = 0 ;j < k ; j++)
                for(g = j+1 ; g < k ; g++)
                    for(e = g+1 ; e < k ; e++)
                    {
                        Point p1 = ed[i][j],p2 = ed[i][g],p3 = ed[i][e];
//                        cout<<p1.x<<" "<<p1.y<<" "<<p2.x<<" "<<p2.y<<" "<<p3.x<<" "<<p3.y<<endl;
//                        cout<<i<<" "<<dis(p1-p2)<<" "<<dis(p1-p3)<<" "<<dis(p2-p3)<<endl;
                        if(dcmp(dis(p1-p2)-dis(p1-p3))==0&&dcmp(dis(p1-p2)-dis(p2-p3))==0&&dcmp(mul(p1,p2,p3))!=0)
                        {
                            sr[cnt++] = i+'a';
                        }
                    }
        }
        if(cnt==0)
        {
            puts("LOOOOOOOOSER!");
            continue;
        }
        for(i = 0 ; i < cnt ; i++)
        printf("%c",sr[i]);
        puts("");
    }
    return 0;
}