Poj 2398 Toy Storage (叉积+二分)-蒲公英云

Poj 2398 Toy Storage (叉积+二分)

第一道用codeblocks敲的题，在换行缩进等方面还不是很适应，居然没有变量名整体替换功能。。。。

变量名字长些可以用查找+替换，短了基本无解。。。

这一题和上一题基本一样，只不过板子没有按照顺序给出，需要排序。最后要求输出包含同样个数的玩具的格子各有多少个

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=1005;
int n,m,x1,y11,x2,y2;
int cnt[N],ans[N];
template<typename Type>
class Point
{
public:
    Type x,y;
    Point(){}
    Point (Type _x,Type _y)
    {
        x=_x;
        y=_y;
    }
    Point operator-(const Point &b) const
    {
        return Point(x-b.x,y-b.y);
    }
    //调用点a的该函数
    //返回正值点a在向量bc的左侧
    //返回负值点a在向量bc的右侧
    //返回0点a在向量bc这条直线上
    Type Cross (Point b,Point c)
    {return (b.x-x)*(c.y-y)-(c.x-x)*(b.y-y);}
};
struct Line
{
    Point<int> up,down;
    bool operator < (const Line& b) const
    {
        return up.x<b.up.x;
    }
}line[N];
Point<int> data[N];
void Init ()
{
    int i,a,b;
    memset(cnt,0,sizeof(cnt));
    memset(ans,0,sizeof(ans));
    for (i=0;i<n;i++)
    {
        scanf("%d%d",&a,&b);
        line[i].up=Point<int>(a,y11);
        line[i].down=Point<int>(b,y2);
    }
    sort(line,line+n);     //此题板子需要排序
    for (i=1;i<=m;i++)
    {
        scanf("%d%d",&a,&b);
        data[i]=Point<int>(a,b);
    }
}
int Solve (Point<int> cc)
{
    int low=0,high=n,mid;
    while (low<=high)
    {
        mid=(low+high)>>1;
        if (line[mid].down.Cross(line[mid].up,cc)>0)
            high=mid-1;
        else
            low=mid+1;
    }
    if (low>n) low=n;
    return low;
}
int main ()
{
#ifdef ONLINE_JUDGE
#else
    freopen("read.txt","r",stdin);
#endif
    while (scanf("%d",&n),n)
    {
        scanf("%d%d%d%d%d",&m,&x1,&y11,&x2,&y2);
        int i;
        Init ();
        for (i=1;i<=m;i++)
            cnt[Solve(data[i])]++;
        printf("Box\n");
        for (i=0;i<=n;i++)
            ans[cnt[i]]++;
        for (i=1;i<=n;i++)
            if (ans[i])
                printf("%d: %d\n",i,ans[i]);
    }
    return 0;
}