POJ 2318 (叉积) TOYS

题意：

有一个长方形，里面从左到右有n条线段，将矩形分成n+1个格子，编号从左到右为0~n。

端点分别在矩形的上下两条边上，这n条线段互不相交。

现在已知m个点，统计每个格子中点的个数。

分析：

用叉积判断点与线段的相对位置，对于每个点二分查找所在的格子。

 #include <cstdio>
 #include <cmath>
 #include <cstring>

 struct Point
 {
     int x, y;
     Point(int x=, int y=):x(x), y(y) {}
 };
 typedef Point Vector;

 Point read_point()
 {
     int x, y;
     scanf("%d%d", &x, &y);
     return Point(x, y);
 }

 Point operator + (const Point& A, const Point& B)
 { return Point(A.x+B.x, A.y+B.y); }

 Point operator - (const Point& A, const Point& B)
 { return Point(A.x-B.x, A.y-B.y); }

 int Cross(const Point& A, const Point& B)
 { return A.x*B.y - A.y*B.x; }

 const int maxn =  + ;
 int up[maxn], down[maxn], ans[maxn];
 int n, m, kase = ;
 Point A0, B0;

 int binary_search(const Point& P)
 {
     int L = , R = n;
     while(L < R)
     {
         int mid = L + (R - L + ) / ;
         Point A(down[mid], B0.y), B(up[mid], A0.y);
         Vector v1 = B - A;
         Vector v2 = P - A;
         if(Cross(v1, v2) < ) L = mid;
         else R = mid - ;
     }
     return L;
 }

 int main()
 {
     //freopen("in.txt", "r", stdin);

     while(scanf("%d", &n) ==  && n)
     {
         memset(ans, , sizeof(ans));

         scanf("%d", &m); A0 = read_point(); B0 = read_point();
         for(int i = ; i <= n; ++i) scanf("%d%d", &up[i], &down[i]);
         for(int i = ; i < m; ++i)
         {
             Point P; P = read_point();
             int pos = binary_search(P);
             ans[pos]++;
         }

         if(kase++) puts("");
         for(int i = ; i <= n; ++i) printf("%d: %d\n", i, ans[i]);
     }

     return ;
 }

代码君