POJ2318 TOYS

嘟嘟嘟

题面：先告诉你一个矩形抽屉的坐标，然后\(n\)个隔板将抽屉分成了\(n + 1\)格（格子从\(0\)到\(n - 1\)标号），接下来随机输入\(m\)个玩具的坐标。问最后每一个格子里有多少个玩具。

仔细想想就是一道计算几何入门题。
对于一个玩具\((x_0, y_0)\)，我们只要找到在他的左面且离他最近的隔板\(i\)，则这个玩具就在第\(i\)格里。
怎么判断隔板\(AB\)（规定\(A\)点是隔板的上端，\(B\)点是隔板下端）是否在玩具的左边？用叉积即可：只要\(\overrightarrow{OA} \times \overrightarrow{OB} > 0\)，就说明在左边。
然后因为有人说从左到右枚举会超时，所以我就把枚举改成倍增了。
然鹅枚举\(O(n ^2)\)能过呀，于是我就枚举交了一发，也过了……（？？？）
代码里注释掉的是枚举的

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 5e3 + 5;
inline ll read()
{ll ans = 0;char ch = getchar(), last = ' ';while(!isdigit(ch)) {last = ch; ch = getchar();}while(isdigit(ch)) {ans = (ans << 1) + (ans << 3) + ch - '0'; ch = getchar();}if(last == '-') ans = -ans;return ans;
}
inline void write(ll x)
{if(x < 0) x = -x, putchar('-');if(x >= 10) write(x / 10);putchar(x % 10 + '0');
}int n, m, xl, yl, xr, yr; 
int ans[maxn];struct Vec
{int x, y;int operator * (const Vec& oth)const{return x * oth.y - oth.x * y;}
};
struct Point
{int x, y;Vec operator - (const Point& oth)const{return (Vec){x - oth.x, y - oth.y};}
}ad[maxn], au[maxn], P;const int N = 13;
void solve()
{int L = 0;for(int i = N, t; i >= 0; --i) if((t = L + (1 << i)) <= n){if((au[t] - P) * (ad[t] - P) > 0) L = t;}/*int L = 0;while(L < n){if((au[L + 1] - P) * (ad[L + 1] - P) < 0) break;L++;}*/ans[L]++;
}int main()
{while(scanf("%d", &n) && n){Mem(ans, 0);m = read();xl = read(); yl = read(); xr = read(); yr = read();au[0].y = yl;for(int i = 1; i <= n; ++i) {au[i].x = read(); au[i].y = yl;ad[i].x = read(); ad[i].y = yr;}for(int i = 1; i <= m; ++i){P.x = read(); P.y = read();solve();}for(int i = 0; i <= n; ++i) write(i), putchar(':'), space, write(ans[i]), enter;enter;}return 0;
}