HDU - 4082 Hou Yi's secret

题意：射箭落在n个点，任取三点可构成一个三角形，问最大的相似三角形集（一组互相相似的三角形）的个数。

分析：

1、若n个点中有相同的点，要去重，题目中说射箭会形成洞，任选三个洞构成三角形，因此射在同一点只形成一个洞。

2、二进制枚举子集选出三个点，判断能否构成三角形。

3、因为边长可能为小数，因此用边长的平方判断相似。

(1)将比较的两个三角形三边分别排序

(2)tmpx[0] / tmpy[0] = tmpx[1] / tmpy[1] = tmpx[2] / tmpy[2]，即若满足tmpx[0] * tmpy[1] == tmpy[0] * tmpx[1] && tmpx[1] * tmpy[2] == tmpx[2] * tmpy[1]，则两三角形相似。

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<cmath>

#include<iostream>

#include<sstream>

#include<iterator>

#include<algorithm>

#include<string>

#include<vector>

#include<set>

#include<map>

#include<stack>

#include<deque>

#include<queue>

#include<list>

#define lowbit(x) (x & (-x))

const double eps = 1e-8;

inline int dcmp(double a, double b){

    if(fabs(a - b) < eps) return 0;

    return a > b ? 1 : -1;

}

typedef long long LL;

typedef unsigned long long ULL;

const int INT_INF = 0x3f3f3f3f;

const int INT_M_INF = 0x7f7f7f7f;

const LL LL_INF = 0x3f3f3f3f3f3f3f3f;

const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;

const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};

const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};

const LL MOD = 1e9 + 7;

const double pi = acos(-1.0);

const int MAXN = 1000 + 10;

const int MAXT = 1000000 + 10;

using namespace std;

struct Node{

    int x, y;

}num[20];

struct tri{

    LL a, b, c;

}num1[MAXN];

int cnt;

int cntpoint;

vector<int> v;

LL tmpx[5];

LL tmpy[5];

map<pair<int, int>, int> mp;

bool judge(LL x, LL y){

    tmpx[0] = num1[x].a;

    tmpx[1] = num1[x].b;

    tmpx[2] = num1[x].c;

    tmpy[0] = num1[y].a;

    tmpy[1] = num1[y].b;

    tmpy[2] = num1[y].c;

    sort(tmpx, tmpx + 3);

    sort(tmpy, tmpy + 3);

    if(tmpx[0] * tmpy[1] == tmpy[0] * tmpx[1] && tmpx[1] * tmpy[2] == tmpx[2] * tmpy[1]) return true;

    return false;

}

int main(){

    int n;

    while(scanf("%d", &n) == 1){

        if(!n) return 0;

        mp.clear();

        cntpoint = 0;

        int x, y;

        for(int i = 0; i < n; ++i){

            scanf("%d%d", &x, &y);

            if(!mp.count(pair<int, int>(x, y))){

                mp[pair<int, int>(x, y)] = 1;

                num[cntpoint].x = x;

                num[cntpoint++].y = y;

            }

        }

        cnt = 0;

        for(int i = 0; i < (1 << cntpoint); ++i){

            v.clear();

            for(int j = 0; j < cntpoint; ++j){

                if(i & (1 << j)){

                    v.push_back(j);

                }

            }

            if(v.size() == 3){

                int tmp11 = (num[v[0]].x - num[v[1]].x) * (num[v[0]].x - num[v[1]].x) +(num[v[0]].y - num[v[1]].y) * (num[v[0]].y - num[v[1]].y);

                int tmp22 = (num[v[1]].x - num[v[2]].x) * (num[v[1]].x - num[v[2]].x) +(num[v[1]].y - num[v[2]].y) * (num[v[1]].y - num[v[2]].y);

                int tmp33 = (num[v[2]].x - num[v[0]].x) * (num[v[2]].x - num[v[0]].x) +(num[v[2]].y - num[v[0]].y) * (num[v[2]].y - num[v[0]].y);

                double tmp1 = sqrt(double(tmp11));

                double tmp2 = sqrt(double(tmp22));

                double tmp3 = sqrt(double(tmp33));

                if(dcmp(tmp1 + tmp2, tmp3) > 0 && dcmp(tmp2 + tmp3, tmp1) > 0 && dcmp(tmp1 + tmp3, tmp2) > 0){

                    num1[cnt].a = (LL)tmp11;

                    num1[cnt].b = (LL)tmp22;

                    num1[cnt].c = (LL)tmp33;

                    ++cnt;

                }

            }

        }

        int ans = 0;

        for(int i = 0; i < cnt; ++i){

            int sum = 1;

            for(int j = 0; j < cnt; ++j){

                if(i != j){

                    if(judge(i, j)) ++sum;

                }

            }

            ans = max(ans, sum);

        }

        printf("%d\n", ans);

    }

    return 0;

}