题目来源:http://poj.org/problem?id=2318
一个 矩阵 被分成多个 区域, 然后输入 多个点, 输出 每个区域点的 个数。 当寻找点 落在某个区域时, 用二分法。
#include <cstdlib> #include <cstring> #include <algorithm> #include <cstdio> #include <cmath> #include <iostream> #include <vector> using namespace std; typedef long long ll; const int N =6000; const double PI = 3.1415927; struct Point{ int x,y; Point(){} Point(int x,int y):x(x),y(y){} // 构造函数,方便代码编写 Point(const Point & p):x(p.x),y(p.y){} Point operator +(Point p){ return Point(x+p.x,y+p.y); } Point operator-(Point p){ return Point(x-p.x,y-p.y); } Point operator*(double d){ return Point(x*d,y*d); } int operator*(Point p){ // 内积 点乘 return x*p.x+ y*p.y; } int operator^(Point p){// 外积 叉乘 return x*p.y-y*p.x; } friend ostream& operator<<(ostream& os,const Point& p ){ os<<p.x<<" "<<p.y<<endl; return os; } friend istream& operator>>(istream& is, Point& p) {// Point 不能是常量,是变量 is>>p.x>>p.y; return is; } }; Point up[N]; Point low[N]; int num[N]; int n; void bin_search(Point p) // 二分法,寻找点所在的区域,用叉乘,注意结束条件是 left+1== right { int left=0,mid; int right = n+1; while(left<=right) { mid=(left+right) /2; int t=(up[mid]-low[mid])^(p-low[mid]); if( t >0 ) right=mid; else if( t <0 ) left=mid; if(left+1 == right ) { num[left]++; break; } } } int main() { int m; int a,b,count=0; Point p; while(scanf("%d",&n)!=EOF && n) { if(count) printf("\n"); count=1; scanf("%d",&m); scanf("%d%d%d%d",&up[0].x,&up[0].y,&low[n+1].x,&low[n+1].y); low[0].x=up[0].x; up[n+1].x=low[n+1].x; for(int i=0;i<=n+1;i++) { up[i].y=up[0].y; low[i].y=low[n+1].y; } for(int i=1;i<=n;i++) { scanf("%d%d",&a,&b); up[i].x=a; low[i].x=b; } memset(num,0,sizeof(num)); for(int i=1;i<=m;i++) { scanf("%d%d",&p.x,&p.y); bin_search(p); } for(int i=0;i<=n;i++) printf("%d: %d\n",i,num[i]); } return 0; }