fin 邊界 vector nta 鏈接 ++ 輸出 mine puts

題目鏈接：http://poj.org/problem?id=2398

Time Limit: 1000MS Memory Limit: 65536K

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 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

與POJ 2318幾乎一模一樣的題。

POJ 2318的題解：http://www.cnblogs.com/dilthey/p/7767218.html

由於本題輸入cardboard的時候是亂序，所以在二分前需要sort一下；另外輸出的方式和2318不太一樣，改一下即可。

AC代碼：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
#define MAX 5005
#define M_PI 3.14159265358979323846 //POJ的math頭文件好像沒有這個定義
using namespace std;

const double eps = 1e-6;

struct Point{
    double x,y;
    Point(double tx=0,double ty=0):x(tx),y(ty){}
};
typedef Point Vctor;

Vctor operator + (Vctor A,Vctor B){return Vctor(A.x+B.x,A.y+B.y);}
Vctor operator - (Point A,Point B){return Vctor(A.x-B.x,A.y-B.y);}
Vctor operator * (Vctor A,double p){return Vctor(A.x*p,A.y*p);}
Vctor operator / (Vctor A,double p){return Vctor(A.x/p,A.y/p);}
bool operator < (Point A,Point B){return A.x < B.x || (A.x == B.x && A.y < B.y);}

struct Line{
    Point p;
    Vctor v;
    Line(Point p=0,Vctor v=Vctor(0,0)):p(p),v(v){}
    Point point(double t){return p + v*t;} //獲得直線上的距離p點t個單位長度的點
};

double Cross(Vctor A,Vctor B){return A.x*B.y-A.y*B.x;}

int dcmp(double x)
{
    if(fabs(x)<eps) return 0;
    else return (x<0)?(-1):(1);
}
bool operator == (Point A,Point B){return dcmp(A.x-B.x)==0 && dcmp(A.y-B.y)==0;}

int n,m;
Point upper_left,lower_right;
Line x_axis;//下邊界
Line cardboard[MAX];//隔板
Point toy;
int area[MAX],cnt[MAX];//記錄每個區域的

int Left_of_Line(Line l,Point p)
{
    if(Cross(l.v,p-l.p)>0) return 1;//左邊
    else return 0;
}
int where(Point toy)
{
    int l=0,r=n+1;
    while(r-l>1)
    {
        int mid=(l+r)/2;
        if(Left_of_Line(cardboard[mid],toy)) r=mid;
        else l=mid;
    }
    return l;
}

bool cmp(Line a,Line b)
{
    if(a.p==b.p) return (a.p+a.v)<(b.p+b.v);
    else return a.p<b.p;
}
int main()
{
    while(scanf("%d",&n) && n!=0)
    {
        scanf("%d%lf%lf%lf%lf",&m,&upper_left.x,&upper_left.y,&lower_right.x,&lower_right.y);

        x_axis=Line(lower_right,Vctor(-1,0));
        cardboard[0]=Line(Point(upper_left.x,lower_right.y),Vctor(0,1)), cardboard[n+1]=Line(lower_right,Vctor(0,1));

        Point U=Point(0,upper_left.y),L=Point(0,lower_right.y);
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf",&U.x,&L.x);
            cardboard[i]=Line(L,U-L);
        }
        sort(cardboard+1,cardboard+n+1,cmp);

        memset(area,0,sizeof(area));
        for(int i=1;i<=m;i++)
        {
            scanf("%lf%lf",&toy.x,&toy.y);
            area[where(toy)]++;
        }
        memset(cnt,0,sizeof(cnt));
        for(int i=0;i<=n;i++) cnt[area[i]]++;
        printf("Box\n");
        for(int i=1;i<=m;i++)
        {
            if(cnt[i]==0) continue;
            printf("%d: %d\n",i,cnt[i]);
        }
    }
}

POJ 2398 - Toy Storage - [計算幾何基礎題][同POJ2318]