題目鏈接:http://acm.hdu.edu.cn/showproblem.php?pid=6731
題意:n個點,m次查詢,每次給你一個點,從n個點挑出兩個能組成多少直角三角形
題解:假設詢問的是點p,直角可能是p也可能是其他兩點的某個,當是其他點時,我們枚舉n個點,讓他作爲直角,這樣就能確定出一條直線,在直線上的點數就是當前能構成直角三角形的個數,如果在通過枚舉的話肯定是不行的,所以我們預處理出每兩個點直線方程,k和b,哈希保存下來(這個直接用long double就可以了,之前用的用1e9+7來哈希,但是超時了),然後就可以二分確定下當前的數目,如果直角是p這個點,那麼也是相同的處理方法,但是這樣得到的數目要除2,因爲多算了一半,唉...
#include <bits/stdc++.h>
#include <tr1/unordered_map>
using namespace std;
typedef long long ll;
const ll mod = 2e9 + 7;
const ll jz = 100007;
const int N = 2010;
struct node {
ll x, y;
}p[N];
int n, m;
tr1::unordered_map<ll, ll> mp;
vector<long double> v[N];
ll num[N];
vector<long double> vp;
ll nump;
/*
ll ksm(ll x, ll y) {
// if(mp[x]) return mp[x];
ll res = 1;
while(y) {
if(y & 1) res = res * x % mod;
y >>= 1;
x = x * x % mod;
}
return res;
}
*/
int main() {
long double k, b;
ll x, y;
ll res1, res2;
long double cnt;
while(~scanf("%d %d", &n, &m)) {
for(int i = 1; i <= n; i++) {
scanf("%lld %lld", &p[i].x, &p[i].y);
v[i].clear();
num[i] = 0;
}
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
if(i == j) continue;
if(p[i].x == p[j].x) num[i]++;
else {
k = (long double)(p[i].y - p[j].y) / (p[i].x - p[j].x);
b = p[i].y - k * p[i].x;
v[i].push_back(k * mod + b);
}
}
sort(v[i].begin(), v[i].end());
}
while(m--) {
scanf("%lld %lld", &x, &y);
res1 = res2 = 0;
nump = 0;
vp.clear();
for(int i = 1; i <= n; i++) {
if(y == p[i].y) res1 += num[i];
else {
k = (long double)(p[i].x - x) / (y - p[i].y);
b = p[i].y - k * p[i].x;
cnt = k * mod + b;
res1 += upper_bound(v[i].begin(), v[i].end(), cnt) - lower_bound(v[i].begin(), v[i].end(), cnt);
}
if(p[i].x == x) nump++;
else {
k = (long double)(p[i].y - y) / (p[i].x - x);
b = p[i].y - k * p[i].x;
cnt = k * mod + b;
vp.push_back(cnt);
}
}
sort(vp.begin(), vp.end());
// cout << nump << endl;
for(int i = 1; i <= n; i++) {
if(y == p[i].y) res2 += nump;
else {
k = (long double)(p[i].x - x) / (y - p[i].y);
b = y - k * x;
cnt = k * mod + b;
res2 += upper_bound(vp.begin(), vp.end(), cnt) - lower_bound(vp.begin(), vp.end(), cnt);
}
}
// cout << res1 << " " << res2 << endl;
printf("%lld\n", res1 + res2 / 2);
}
}
return 0;
}
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