1 // 题意：问你每个区域有多少个点 2 // 思路：数据小可以直接暴力 3 // 也可以二分区间 4 5 #include <cstdio> 6 #include <cstring> 7 #include <iostream> 8 #include <algorithm> 9 #include <map> 10 #include <set> 11 #include <queue> 12 #include <stdlib.h> 13 #include <cmath> 14 using namespace std; 15 typedef long long LL; 16 const LL inf = 1e18; 17 const int N = 5000; 18 19 struct Point{ 20 int x,y; 21 Point(){} 22 Point(int _x,int _y){ 23 x=_x;y=_y; 24 } 25 Point operator -(const Point &b)const{ 26 return Point(x-b.x,y-b.y); 27 } 28 int operator *(const Point &b)const{ 29 return x*b.x+y*b.y; 30 } 31 int operator ^(const Point &b)const{ 32 return x*b.y-y*b.x; 33 } 34 }; 35 36 struct Line{ 37 Point s,e; 38 Line(){} 39 Line(Point _s,Point _e){ 40 s=_s,e=_e; 41 } 42 }; 43 44 int xmult(Point p0,Point p1,Point p2){ 45 return (p1-p0)^(p2-p0); 46 } 47 48 Line line[N]; 49 int ans[N]; 50 int main(){ 51 int n,m,x1,y1,x2,y2; 52 bool first = true; 53 while(~scanf("%d",&n)&&n){ 54 if(first) first=false; 55 else printf(" "); 56 scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2); 57 for(int i=0;i<n;i++){ 58 int x,y; 59 scanf("%d%d",&x,&y); 60 line[i]=Line(Point(x,y1),Point(y,y2)); 61 } 62 line[n]=Line(Point(x2,y1),Point(x2,y2)); 63 int x,y; 64 Point p; 65 memset(ans,0,sizeof(ans)); 66 int tem; 67 while(m--){ 68 scanf("%d%d",&x,&y); 69 p = Point(x,y); 70 int l=0,r=n,mid; 71 while(l<=r){ 72 mid=(l+r)>>1; 73 if(xmult(p,line[mid].s,line[mid].e)<0){ 74 tem=mid; 75 r=mid-1; 76 } 77 else l=mid+1; 78 } 79 ans[tem]++; 80 } 81 for(int i=0;i<=n;i++) printf("%d: %d ",i,ans[i]); 82 } 83 return 0; 84 }
1 // POJ 2398和2318差不多 多了排序和统计输出 2 3 #include <cstdio> 4 #include <cstring> 5 #include <iostream> 6 #include <algorithm> 7 #include <map> 8 #include <set> 9 #include <queue> 10 #include <stdlib.h> 11 #include <cmath> 12 using namespace std; 13 typedef long long LL; 14 const LL inf = 1e18; 15 const int N = 5000; 16 17 struct Point{ 18 int x,y; 19 Point(){} 20 Point(int _x,int _y){ 21 x=_x;y=_y; 22 } 23 Point operator -(const Point &b)const{ 24 return Point(x-b.x,y-b.y); 25 } 26 int operator *(const Point &b)const{ 27 return x*b.x+y*b.y; 28 } 29 int operator ^(const Point &b)const{ 30 return x*b.y-y*b.x; 31 } 32 }; 33 34 struct Line{ 35 Point s,e; 36 Line(){} 37 Line(Point _s,Point _e){ 38 s=_s,e=_e; 39 } 40 }; 41 42 int xmult(Point p0,Point p1,Point p2){ 43 return (p1-p0)^(p2-p0); 44 } 45 46 bool cmp(Line a,Line b){ 47 return a.s.x<b.s.x; 48 } 49 50 Line line[N]; 51 int ans[N]; 52 int num[N]; 53 int main(){ 54 int n,m,x1,y1,x2,y2; 55 while(~scanf("%d",&n)&&n){ 56 memset(num,0,sizeof(num)); 57 scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2); 58 for(int i=0;i<n;i++){ 59 int x,y; 60 scanf("%d%d",&x,&y); 61 line[i]=Line(Point(x,y1),Point(y,y2)); 62 } 63 line[n]=Line(Point(x2,y1),Point(x2,y2)); 64 sort(line,line+n+1,cmp); 65 int x,y; 66 Point p; 67 memset(ans,0,sizeof(ans)); 68 int tem; 69 while(m--){ 70 scanf("%d%d",&x,&y); 71 p = Point(x,y); 72 int l=0,r=n,mid; 73 while(l<=r){ 74 mid=(l+r)>>1; 75 if(xmult(p,line[mid].s,line[mid].e)<0){ 76 tem=mid; 77 r=mid-1; 78 } 79 else l=mid+1; 80 } 81 ans[tem]++; 82 } 83 for(int i=0;i<=n;i++){ 84 num[ans[i]]++; 85 } 86 printf("Box "); 87 for(int i=1;i<=n;i++) { 88 if(num[i]>0) 89 printf("%d: %d ",i,num[i]); 90 } 91 } 92 return 0; 93 }