TOYS（计算几何基础+点与直线的位置关系）

题目链接：http://poj.org/problem?id=2318

题面：

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　TOYS

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17413		Accepted: 8300

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.

 

思路：对于每一条分界线，首先进行排序（根据他的上面的x排和下面的排都是可以的），然后标号从0到n-1。然后问题就转换为现在访问的点会在哪条分界线的左端，若落在第i条直线的左端，那么它也就落在区域i内，确定所属区域后直接return。在最后面对第n个区域进行处理前面没被return的点。对于本题的关键就是处理点与直线的位置关系，而这个用向量的叉积进行处理即可。

代码实现如下：

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 using namespace std;

 const int maxn = 5e3 + ;
 int n, m, x1, x2, y1, y2;
 int num[maxn];

 struct node {
     int x, y;
     bool operator < (const node &b) const {
         return x < b.x;
     }
 } P[maxn], L[maxn], nw, nxt;  //P储存所访问的点（其实也可以不用数组的），L储存直线（对于直线此处的x为上方的x，y为下方的y）；

 double dot(node a, node b) {
     return (a.x * b.y - a.y * b.x);
 }

 bool check(node p, node pp) {
     nw.x = p.x - pp.y, nw.y = p.y - y2, nxt.x = pp.x - pp.y, nxt.y = y1 - y2;
     if( dot(nw, nxt) < ) {  //为负则说明当前访问的点在该直线的左端；
         return true;
     } else
         return false;
 }

 void Throw(node p) {
     for(int i = ; i < n; i++) {
         if(check(p, L[i])) {
             num[i]++;
             return;
         }
     }
     num[n]++;
     return;
 }

 int main() {
     //freopen("in.txt", "r", stdin);
     while(~scanf("%d", &n) && n) {
         scanf("%d%d%d%d%d", &m, &x1, &y1, &x2, &y2);
         memset(num, , sizeof(num));
         for(int i = ; i < n; i++) {
             scanf("%d%d", &L[i].x, &L[i].y);
         }
         sort(L, L + n);
         for(int i = ; i < m; i++) {
             scanf("%d%d", &P[i].x, &P[i].y);
             Throw(P[i]);
         }
         for(int i = ; i <= n; i++) {
             printf("%d: %d\n", i, num[i]);
         }
         printf("\n");
     }
     return ;
 }

18年10月6日更新：

前面的代码是以前看题解写的，今天再做一次，发现就是求一个叉积的事。

思路：首先将所有直线用结构体存起来，按照u排序，若u相同则按照l排序；这样就使得线段是有序的，第1条线段左边是0号区域，第2条左边是1号……第n条左边是n-1号，右边是n号。将每个点依次与这些直线求叉积，若该点在第i条直线的顺时针方向（也就是叉积为正数），那么点必落在第i-1号区域；若与所有直线的叉积都为负数则落在第n号区域。

代码实现如下：

 #include <set>
 #include <map>
 #include <deque>
 #include <ctime>
 #include <stack>
 #include <cmath>
 #include <queue>
 #include <string>
 #include <cstdio>
 #include <vector>
 #include <iomanip>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long LL;
 typedef pair<LL, LL> pll;
 typedef pair<LL, int> pli;
 typedef pair<int, int> pii;
 typedef unsigned long long uLL;

 #define lson rt<<1
 #define rson rt<<1|1
 #define name2str(name)(#name)
 #define bug printf("**********\n");
 #define IO ios::sync_with_stdio(false);
 #define debug(x) cout<<#x<<"=["<<x<<"]"<<endl;
 #define FIN freopen("/home/dillonh/CLionProjects/in.txt","r",stdin);

 const double eps = 1e-;
 const int maxn =  + ;
 const int inf = 0x3f3f3f3f;
 const double pi = acos(-1.0);
 const LL INF = 0x3f3f3f3f3f3f3f3fLL;

 int n, m, x1, yy, x2, y2, x, y;
 int cnt[maxn];

 struct Line {
     int l, r;
     bool operator < (const Line& x) const {
         return l == x.l ? r < x.r : l < x.l;
     }
 }L[maxn];

 int cross(int x1, int y1, int x2, int y2, int x3, int y3) {
     return (x2 - x1) * (y3 - y1) - (y2 - y1) * (x3 - x1);
 }

 int main() {
 #ifndef ONLINE_JUDGE
     FIN;
 #endif
     int vis = ;
     while(~scanf("%d", &n) && n) {
         if(vis) printf("\n");
         vis = ;
         scanf("%d%d%d%d%d", &m, &x1, &yy, &x2, &y2);
         for(int i = ; i <= n; i++) {
             scanf("%d%d", &L[i].l, &L[i].r);
         }
         sort(L + , L + n + );
         memset(cnt, , sizeof(cnt));
         for(int i = ; i <= m; i++) {
             scanf("%d%d", &x, &y);
             int flag = ;
             for(int j = ; j <= n; j++) {
                 if(cross(L[j].r, y2, L[j].l, yy, x, y) > ) {
                     flag = ;
                     cnt[j-]++;
                     break;
                 }
             }
             if(!flag) cnt[n]++;
         }
         for(int i = ; i <= n; i++) {
             printf("%d: %d\n", i, cnt[i]);
         }
     }
     return ;
 }