poj2318 TOYS

最新推荐文章于 2021-05-12 20:50:00 发布

weixin_30410119

最新推荐文章于 2021-05-12 20:50:00 发布

阅读量37

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/yuyanjiaB/p/9996295.html

传送门

给一个矩形和n块隔板,m个点问隔板隔开的每个区域内有多少个点(保证不在隔板上)

n<=5000,m<=5000

首先考虑如果所有板子都是直的就是一个二分查找的标准形式

其实斜的也能一样做因为二分的条件没变

变得就是如何判断在板子的左侧之类的

一个叉积就好了

图源自dalao https://blog.youkuaiyun.com/yskyskyer123/article/details/52107419

就是Xmlt(PA,PB)<0因为卡格所以没有共线的问题吧

复杂度O(mlgn*浮点常数)

Code:

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<cmath>
  5 #include<queue>
  6 #include<iostream>
  7 #define ms(a,b) memset(a,b,sizeof a)
  8 #define rep(i,a,n) for(int i = a;i <= n;i++)
  9 #define per(i,n,a) for(int i = n;i >= a;i--)
 10 #define inf 2147483647
 11 using namespace std;
 12 typedef long long ll;
 13 typedef double D;
 14 #define eps 1e-8
 15 ll read() {
 16     ll as = 0,fu = 1;
 17     char c = getchar();
 18     while(c < '0' || c > '9') {
 19         if(c == '-') fu = -1;
 20         c = getchar();
 21     }
 22     while(c >= '0' && c <= '9') {
 23         as = as * 10 + c - '0';
 24         c = getchar();
 25     }
 26     return as * fu;
 27 }
 28 //head
 29 #define P point
 30 struct point {
 31     D x,y;
 32     point(){}
 33     point(D X,D Y):x(X),y(Y){}
 34     D len() {return sqrt(x*x+y*y);}
 35     D tan() {return y/x;}
 36     D sin() {return len() / y;}
 37     D cos() {return len() / x;}
 38     void read() {scanf("%lf%lf",&x,&y);}
 39     void print() {printf("%.2lf %.2lf\n",x,y);}
 40     friend inline point operator + (const point &a,const point &b) {
 41         return point(a.x+b.x,a.y+b.y);
 42     }
 43     friend inline point operator - (const point &a,const point &b) {
 44         return point(a.x-b.x,a.y-b.y);
 45     }
 46     //放缩
 47     friend inline point operator * (const point &a,const D &b) {
 48         return point(a.x*b,a.y*b);
 49     }
 50     //叉积
 51     friend inline D operator * (const point &a,const point &b) {
 52         return a.x*b.y-a.y*b.x;
 53     }
 54     //点积
 55     friend inline D operator / (const point &a,const point &b) {
 56         return a.x*b.x+a.y*b.y;
 57     }
 58 };
 59 struct line {
 60     D k,b;
 61     void init(point x,point y) {
 62         k = (y.y-x.y)/(y.x-x.x);
 63         b = x.y - k * x.x;
 64     }
 65     D YY(D X) {return k*X+b;}
 66     D XX(D Y) {return (Y-b)/k;}
 67 };
 68 struct yuan {
 69     D r,x,y;
 70     yuan(){}
 71     yuan(int R,int X,int Y):r(R),x(X),y(Y){}
 72 };
 73 bool ONSEG(point a,point b,point p) {
 74     return ((a-b).len() == (a-p).len() + (p-b).len());
 75 }
 76 D TRIAREA(point a,point b,point c) {
 77     return ((a-b)*(a-c)) / 2.0;
 78 }
 79 #define ONLINE(a,b,c) (((a-b)*(a-c)) == 0)
 80 #define sign(x) (x) > 0 ? 1 : ((x) < 0 ? -1 : 0)
 81 // 1  0 -1
 82 // 锐 直 钝
 83 int ANGDIR(point a,point b,point p) {
 84     D ans = (p-a)*(p-b);
 85     return sign(ans);
 86 }
 87 
 88 D dis(point a,point b,point p) {
 89     if(ANGDIR(b,p,a) == -1) return (p-a).len();
 90     if(ANGDIR(a,p,b) == -1) return (p-b).len();
 91     return ((p-a)*(p-b)) / (a-b).len();
 92 }
 93 D dis(point a,line l) {
 94     return (l.k * a.x - a.y + l.b) / sqrt(l.k*l.k+1);
 95 }
 96 int cross(P a,P b,P c,P d) {
 97     if(ONLINE(a,b,c) ^ ONLINE(a,b,d)) return 1;
 98     if(ONLINE(c,d,a) ^ ONLINE(c,d,b)) return 1;
 99     if(ONLINE(a,b,c) & ONLINE(a,b,d)) return -1;
100     if(ONLINE(c,d,a) & ONLINE(c,d,b)) return -1;
101     D J1 = ((c-d)*(c-a)) * ((c-d)*(c-b));
102     D J2 = ((a-b)*(a-c)) * ((a-b)*(a-d));
103     if(J1 < 0 && J2 < 0) return 1;
104     return 0;
105 }
106 point Cross(point a,point b,point c,point d) {
107     if(ONLINE(a,b,c)) return c;
108     if(ONLINE(a,b,d)) return d;
109     if(ONLINE(c,d,a)) return a;
110     if(ONLINE(c,d,b)) return b;
111     D S1 = (a-c)*(a-d),S2 = (b-d) * (b-c);
112     point tmp = (b-a) * (S1 / (S1+S2));
113     return a + tmp;
114 }
115 //CP
116 const int N = 100003;
117 int n,T;
118 D X1,Y1,X2,Y2;
119 int cnt[N];
120 struct seg {
121     point a,b;
122 }p[N];
123 point aim;
124 bool check(int i) {
125     return (aim - p[i].a) * (aim - p[i].b) < 0;
126 }
127 
128 void solve() {
129     aim.read();
130     int L = 0,R = n+1;
131     while(R > L) {
132         int m = L+R >> 1;
133         if(check(m)) R = m;
134         else L = m+1;
135     }
136     cnt[L]++;
137 }
138 
139 int main() {
140     while(1) {
141         n = read();
142         if(!n) return 0;
143         T = read();
144         X1 = read(),Y1 = read(),X2 = read(),Y2 = read();
145         rep(i,1,n) {
146             p[i].a = point(read(),Y1);
147             p[i].b = point(read(),Y2);
148         }
149         ms(cnt,0);
150         rep(i,1,T) solve();
151         rep(i,1,n+1) printf("%d: %d\n",i-1,cnt[i]);
152         puts("");
153     }
154 }