[POJ 2318]TOYS
阿新 • • 發佈：2017-07-30
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Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toys in, but John 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 John to find his favorite toys.
John‘s parents came up with the following idea. They put cardboard partitions into the box. Even if John keeps throwing his toys into the box, at least toys that get thrown into different bins stay separated. The following diagram shows a top view of an example toy box.
For this problem, you are asked to determine how many toys fall into each partition as John throws them into the toy box. HINT
As the example illustrates, toys that fall on the boundary of the box are "in" the box.
Description
Calculate the number of toys that land in each bin of a partitioned toy box.Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toys in, but John 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 John to find his favorite toys.
John‘s parents came up with the following idea. They put cardboard partitions into the box. Even if John keeps throwing his toys into the box, at least toys that get thrown into different bins stay separated. The following diagram shows a top view of an example toy box.
For this problem, you are asked to determine how many toys fall into each partition as John throws them into the toy box.
Input
The input file contains one or more problems. The first line of a problem consists of six integers, n m x1 y1 x2 y2. The number of cardboard partitions is n (0 < n <= 5000) and the number of toys is m (0 < m <= 5000). 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 contain two integers per line, Ui Li, indicating that the ends of the i-th cardboard partition is at the coordinates (Ui,y1) and (Li,y2). You may assume that the cardboard partitions do not intersect each other and that they are specified in sorted order from left to right. The next m lines contain two integers per line, Xj Yj specifying where the j-th toy has landed in the box. The order of the toy locations is random. You may assume that no toy will land exactly on a cardboard partition or outside the boundary of the box. The input is terminated by a line consisting of a single 0.Output
The output for each problem will be one line for each separate bin in the toy box. For each bin, print its bin number, followed by a colon and one space, followed by the number of toys thrown into that bin. Bins are numbered from 0 (the leftmost bin) to n (the rightmost bin). Separate the output of different problems by a single blank line.Sample Input
5 6 0 10 60 0
3 1
4 3
6 8
10 10
15 30
1 5
2 1
2 8
5 5
40 10
7 9
4 10 0 10 100 0
20 20
40 40
60 60
80 80
5 10
15 10
25 10
35 10
45 10
55 10
65 10
75 10
85 10
95 10
0
Sample Output
0: 2
1: 1
2: 1
3: 1
4: 0
5: 1
0: 2
1: 2
2: 2
3: 2
4: 2
HINT
As the example illustrates, toys that fall on the boundary of the box are "in" the box.
題目大意
給出矩形的左上、右下坐標，按順序給出n條線段在矩形上兩條橫邊上的截距。這n條直線將矩形分成n+1塊。給出m個點，問每個小塊中有幾個點。
題解
還記得之前的某篇博客講過叉積的性質嗎？
叉積的一個非常重要性質是可以通過它的符號判斷兩矢量相互之間的順逆時針關系：
- 若 P × Q > 0 , 則P在Q的順時針方向。
- 若 P × Q < 0 , 則P在Q的逆時針方向。
- 若 P × Q = 0 , 則P與Q共線，但可能同向也可能反向。
這裏同理，我們只需要掃過每條直線判斷點在線的左端還是右端。
當然可以二分實現：我們設判斷的點為(x,y)，直線的上下截距分別為u，d，矩形上邊縱坐標ya，下邊yb。
對於二分出的mid值
if (Point(x-d[mid],y-yb)*Point(u[mid]-d[mid],ya-yb)>0) l=mid+1;
else r=mid-1,ans=mid;
這裏直線0～n-1編號，所以ans初值應該賦為n。
1 #include<set> 2 #include<map> 3 #include<ctime> 4 #include<cmath> 5 #include<queue> 6 #include<stack> 7 #include<cstdio> 8 #include<string> 9 #include<vector> 10 #include<cstring> 11 #include<cstdlib> 12 #include<iostream> 13 #include<algorithm> 14 #define LL long long 15 #define RE register 16 #define IL inline 17 using namespace std; 18 const int N=5000; 19 20 int n,m,xa,xb,ya,yb; 21 int d[N+5],u[N+5]; 22 struct Point 23 { 24 int x,y; 25 Point (){}; 26 Point (int _x,int _y) {x=_x,y=_y;} 27 int operator *(const Point &a) 28 const{ 29 return x*a.y-y*a.x; 30 } 31 }; 32 int cnt[N+5],x,y; 33 34 IL int Dev(int x,int y); 35 36 int main() 37 { 38 scanf("%d",&n); 39 while (n) 40 { 41 memset(cnt,0,sizeof(cnt)); 42 scanf("%d%d%d%d%d",&m,&xa,&ya,&xb,&yb); 43 for (RE int i=0;i<n;i++) scanf("%d%d",&u[i],&d[i]); 44 for (RE int i=1;i<=m;i++) 45 { 46 scanf("%d%d",&x,&y); 47 cnt[Dev(x,y)]++; 48 } 49 for (RE int i=0;i<=n;i++) printf("%d: %d\n",i,cnt[i]); 50 printf("\n"); 51 scanf("%d",&n); 52 } 53 return 0; 54 } 55 56 IL int Dev(int x,int y) 57 { 58 int l=0,r=n-1,ans=n,mid; 59 while (l<=r) 60 { 61 mid=(l+r)>>1; 62 if (Point(x-d[mid],y-yb)*Point(u[mid]-d[mid],ya-yb)>0) l=mid+1; 63 else r=mid-1,ans=mid; 64 } 65 return ans; 66 }
[POJ 2318]TOYS