POJ 2398--Toy Storage(叉积判断，二分找点，点排序)

 1 #include<iostream>
 2 #include<cstring> 
 3 #include<string>
 4 #include<algorithm> 
 5 #include<cstdio> 
 6 #include<cstdlib> 
 7 #include<cmath> 
 8 using namespace std;
 9 const int MAX = 5005;
10 typedef struct point {
11     int x;
12     int y;
13 }point;
14 typedef struct value {
15     point start;
16     point end;
17 }v;
18 v edge[MAX];
19 int sum[MAX], ans[MAX];
20 int n, m, x1, y11, x2, y2, flag = 1,Ui, Li;
21 point tp;
22 int Xj, Yj;
23 bool com(const v t1, const v t2) {
24     return t1.start.x < t2.start.x;
25 }
26 bool com2(const int a, const int b) {
27     return a < b;
28 }
29 int multi(point p1, point p2, point p0) {  //判断p1p0和p2p0的关系，<0,p1p0在p2p0的逆时针方向
30     return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y);
31 }
32 void inset(point p) {
33     int low = 0, high = n;
34     while (low <= high) {
35         int mid = (high + low) / 2;
36         if (multi(p, edge[mid].start, edge[mid].end) < 0)    /*点p1在边的左侧*/
37             high = mid - 1;
38         else    //点p在边的右侧
39             low = mid + 1;
40     }
41     if (multi(p, edge[low-1].start, edge[low-1].end) < 0 )
42         sum[low-1]++;
43     else
44         sum[low]++;
45 }
46 int main() {
47     while (cin>>n && n) {
48         memset(sum, 0, sizeof(sum));
49         memset(ans, 0, sizeof(ans));
50         cin >> m >> x1 >> y11 >> x2 >> y2;
51         for (int i = 0; i < n; i++) {
52             cin >> Ui >> Li;
53             edge[i].start.x = Ui;
54             edge[i].start.y = y11;
55             edge[i].end.x = Li;
56             edge[i].end.y = y2;
57         }
58         edge[n].start.x = x2;
59         edge[n].start.y = y11;
60         edge[n].end.x = x2;
61         edge[n].end.y = y2;
62         sort(edge, edge + n + 1, com);
63         for (int j = 0; j < m; j++) {
64             cin >> Xj >> Yj;
65             tp.x = Xj;
66             tp.y = Yj;
67             inset(tp);
68         }
69         for (int i = 0; i <= n; i++)
70         {
71             if (sum[i] != 0)
72                 ans[sum[i]]++;
73         }
74         cout << "Box" << endl;
75         for (int i = 0; i <= n; i++)
76         {
77             if (ans[i] != 0)
78                 cout << i << ": " << ans[i] << endl;
79         }    
80     }
81     return 0;
82 }

1 再熟悉一下叉积函数
2 int multi(point p1, point p2, point p0) { 
3     return (p1.x - p0.x)*(p2.y - p0.y) - (p2.x - p0.x)*(p1.y - p0.y);
4 }
5  //判断p1p0和p2p0的关系
6  //结果<0, p1p0在p2p0的逆时针方向，即点p1在p2p0的左侧
7  //结果>0, p1p0在p2p0的顺时针方向，即点p1在p2p0的右侧