題干：

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 80 805 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: 10: 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：

? ? 在一個盒子內用擋板隔開給出擋板兩端坐標和一些玩具的坐標輸出落在每個方格中的玩具數量。

解題報告：

? ? ? 通過叉積直接判斷玩具點和直線的位置關系，從而確定玩具歸屬于哪一個格子，二分找位置，找到之后還要來一個if判斷來確認這個位置是否正確（即做±1的微調）。再就是注意格式，輸出之間需要一個空行。

AC代碼：

#include<cstdio> #include<cstring> #include<iostream> #include<algorithm>using namespace std; int n,m,x1,x2,y1,y2; int ans[5005]; int top; struct Point {int x,y;Point(){}Point(int x,int y):x(x),y(y){} } p[5005]; struct Edge {Point s,e;Edge(){}Edge(Point s,Point e):s(s),e(e){} } e[5005];int xmult(Point o,Point a,Point b) {return (a.x-o.x) * (b.y-o.y) - (a.y-o.y) * (b.x-o.x); } int main() {int xx1,xx2,xx,yy;while(~scanf("%d%d%d%d%d%d",&n,&m,&x1,&y1,&x2,&y2)) {if(n == 0) break;memset(ans,0,sizeof(ans));top = 0;for(int i = 1; i<=n; i++) {scanf("%d%d",&xx1,&xx2);e[i] = Edge(Point(xx1,y1),Point(xx2,y2));}e[n+1] = Edge(Point(x2,y1),Point(x2,y2)); for(int i = 1; i<=m; i++) {scanf("%d%d",&xx,&yy);p[i] = Point(xx,yy);}int l,r,mid;for(int i = 1; i<=m; i++) {l = 1; r = n+1;//如果這里是r=n+1，那就必須加 35行的e[n+1] = Edge(Point(x2,y1),Point(x2,y2)); 這一句。但是這里也可以直接r=n，這樣就不需要35行那句了并且也可以ac。想想為什么 mid = (l+r)/2;while(l<r) {mid = (l+r)/2;if(xmult(p[i],e[mid].s,e[mid].e) < 0) r = mid;else l = mid+1;}if(xmult(p[i],e[l].s,e[l].e) < 0) ans[l]++;else ans[l+1]++;}for(int i = 1; i<=n+1; i++) {printf("%d: %d\n",i-1,ans[i]);}puts(""); }return 0 ; }

今天又寫了一遍：（也算是體會到判斷ans[l]++或者ans[l+1]++的真正原因）

#include<cstdio> #include<algorithm> #include<iostream> #include<cstring> using namespace std; int n,m,x1,x2,y1,y2;//n個板子，m個玩具，左上角，右下角。 struct Edge {int x1,y1;//上 int x2,y2;//下 } e[5005]; struct Node {int x,y; } node[5005]; int ans[5005]; int cal(int tar,int line) {return (node[tar].x - e[line].x2)*(e[line].y1-e[line].y2) - (e[line].x1 - e[line].x2)*(node[tar].y - e[line].y2); } int main() {while(cin>>n) {if(n == 0) break;cin>>m>>x1>>y1>>x2>>y2;memset(ans,0,sizeof ans);for(int i = 1; i<=n; i++) {scanf("%d%d",&e[i].x1,&e[i].x2);e[i].y1 = y1;e[i].y2 = y2;}for(int i = 1; i<=m; i++) {scanf("%d%d",&node[i].x,&node[i].y);}//process each for(int i = 1; i<=m; i++) {int l = 1,r = n;int mid = (l+r)/2;while(l<r) {mid = (l+r)/2;if(cal(i,mid) > 0) l=mid+1;else r=mid; }if(l == n) {if(cal(i,l) < 0) ans[l]++;else ans[l+1]++; }else ans[l]++;}for(int i = 1; i<=n+1; i++) {printf("%d: %d\n",i-1,ans[i]);}printf("\n");}return 0 ;}

?

總結

以上是生活随笔為你收集整理的【POJ - 2318】TOYS（计算几何，叉积判断点与直线位置关系，二分）的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。