poj 2398 Toy Storage （计算几何）

Toy Storage

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5487 Accepted: 3268

Description

Mom 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 the box. 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.

Input

The 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-left corner 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 that the 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.

Output

For 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 containing t toys. Output will be sorted in ascending order of t for each box.

Sample Input

4 10 0 10 100 020 2080 8060 6040 405 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

Sample Output

Box2: 5Box1: 42: 1

Source

Tehran 2003 Preliminary

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题目大意：给出一个矩形，被一些线段分成很多四边形，然后判断某个点在哪个四边形中，最后输出包t个点的四边形的个数

题解：同poj 2318 ，用叉积判断点与直线的位置关系。

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<cmath>#define LL long long #define N 1003using namespace std;int n,m,x,y,x2,y2,num[N],ans[N];struct vector {int x,y;vector (int X=0,int Y=0){x=X; y=Y;}}line[N];vector operator -(vector a,vector b) {return vector (a.x-b.x,a.y-b.y);}typedef vector data;data a[N],point[N];int cmp(data a,data b){return a.x<b.x||a.x==b.x&&a.y<b.y;}int cross(vector a,vector b){return a.x*b.y-a.y*b.x;}int main(){freopen("a.in","r",stdin);while (true) {scanf("%d",&n);if (!n) break;scanf("%d%d%d%d%d",&m,&x,&y,&x2,&y2);a[1].x=x; a[1].y=x;for (int i=2;i<=n+1;i++)  scanf("%d%d",&a[i].x,&a[i].y);sort(a+1,a+n+1+1,cmp);a[n+2].x=x2; a[n+2].x=x2;line[1].x=x-x; line[1].y=y-y2;for (int i=2;i<=n+1;i++) line[i].x=a[i].x-a[i].y,line[i].y=y-y2;line[n+2]=x2-x2; line[n+2].y=y-y2;for (int i=2;i<=n+1;i++) point[i].x=a[i].y,point[i].y=y2;point[1].x=x; point[1].y=y2;point[n+2].x=x2; point[n+2].y=y2;     memset(num,0,sizeof(num));    for (int i=1;i<=m;i++) {       data now; scanf("%d%d",&now.x,&now.y);       for (int j=1;j<=n+1;j++) {         vector t=now-point[j]; vector t1=now-point[j+1];         if (cross(line[j],t)<0&&cross(line[j+1],t1)>0) { num[j]++; break;  }   }}memset(ans,0,sizeof(ans));for (int i=1;i<=n+2;i++) ans[num[i]]++; printf("Box\n");for (int i=1;i<=m;i++) if (ans[i]) printf("%d: %d\n",i,ans[i]); }}