POJ 2398 Toy Storage 叉积 和2318基本一样

B - Toy StorageTime Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d& %I64uSubmit Status Practice POJ2398Appoint description: System Crawler  (2016-05-10)DescriptionMom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza is rebellious and obeys his parents by simply throwing his toys into thebox. All the toys get mixed up, and it is impossible for Reza to find his favorite toys anymore. Reza's parents came up with the following idea. They put cardboard partitions into the box. Even if Reza keeps throwing his toys into the box, at least toys that get thrown into different partitions stay separate. The box looks like this from the top: We want for each positive integer t, such that there exists a partition with t toys, determine how many partitions have t, toys.InputThe input consists of a number of cases. The first line consists of six integers n, m, x1, y1, x2, y2. The number of cardboards to form the partitions is n (0 < n <= 1000) and the number of toys is given in m (0 < m <= 1000). The coordinates of the upper-leftcorner and the lower-right corner of the box are (x1, y1) and (x2, y2), respectively. The following n lines each consists of two integers Ui Li, indicating that the ends of the ith cardboard is at the coordinates (Ui, y1) and (Li, y2). You may assume thatthe cardboards do not intersect with each other. The next m lines each consists of two integers Xi Yi specifying where the ith toy has landed in the box. You may assume that no toy will land on a cardboard. A line consisting of a single 0 terminates the input.OutputFor each box, first provide a header stating "Box" on a line of its own. After that, there will be one line of output per count (t > 0) of toys in a partition. The value t will be followed by a colon and a space, followed the number of partitions containingt toys. Output will be sorted in ascending order of t for each box.Sample Input

4 10 0 10 100 0
20 20
80 80
60 60
40 40
5 10
15 10
95 10
25 10
65 10
75 10
35 10
45 10
55 10
85 10
5 6 0 10 60 0
4 3
15 30
3 1
6 8
10 10
2 1
2 8
1 5
5 5
40 10
7 9
0

Sample Output

Box
2: 5
Box
1: 4
2: 1

在poj2138的基础上对线段排序就ok

ACcode：

#include <cstring>#include <algorithm>#include <cstdio>#define maxn 5050struct  Point{int x,y;Point (){}Point(int _x,int _y){x=_x;y=_y;}bool operator<(Point b)const{return (x-b.x==0)?y-b.y:x<b.x;}bool operator==(Point b)const{return (x-b.x==0)&&(y-b.y==0);}}p[maxn];struct Line{Point s,e;Line(){}Line(Point _s,Point _e){s=_s;e=_e;}bool operator <(Line t)const{return (s==t.s)?e<t.e:s<t.s;}}my[maxn];int ans[maxn];int multi(Point p1,Point  p2,Point p0){return (p1.x-p0.x)*(p2.y-p0.y)-(p2.x-p0.x)*(p1.y-p0.y);}int main(){int n,m,x1,y1,x2,y2;while(~scanf("%d",&n),n){scanf("%d%d%d%d%d",&m,&x1,&y1,&x2,&y2);for(int i=0;i<n;++i){int a,b;scanf("%d%d",&a,&b);my[i]=Line(Point(a,y1),Point(b,y2));}my=Line(Point(x2,y1),Point(x2,y2));memset(ans,0,sizeof(ans));std::sort(my,my+n+1);for(int i=1;i<=m;++i){int x,y;scanf("%d%d",&x,&y);Point tmp=Point(x,y);int l=0,r=n,id;while(l<=r){int mid=(l+r)>>1;if(multi(my[mid].s,my[mid].e,tmp)<0){id=mid;r=mid-1;}else l=mid+1;}ans[id]++;}std::sort(ans,ans+n+1);puts("Box");int cnt;for(int i=0;i<=n;++i){cnt=1;if(ans[i]==0)continue;while(ans[i]==ans[i+1]){i++;cnt++;}printf("%d: %d\n",ans[i],cnt);}}return 0;}/*4 10 0 10 100 020 2040 4060 6080 805 1015 1095 1025 1065 1075 1035 1045 1055 1085 105 6 0 10 60 04 315 303 16 810 102 12 81 55 540 107 90*/