题目链接：POJ
2398 Toy Storage
之前做的类似题目：POJ 2318 TOYS
【题意】跟之前做的POJ
2318差不多额，给你一个矩形，有被若干直线分成N个格子，给出M个点（玩具）的坐标，问你放有t个玩具的格子的个数。
【思路】其实跟POJ
2318差不多，利用叉积+二分，但是本题中直线的输入不是按顺序的，要进行排序，才能做二分。开始写错了构造函数，然后就一直不对啊，看来C++的学习之路还很远啊。
1 /*
2 ** POJ 2398 Toy Storage
3 ** Created by Rayn @@ 2014/05/05
4 */
5 #include <cstdio>
6 #include <cstring>
7 #include <algorithm>
8 using namespace std;
9 const int MAX = 1010;
10 const int INF = 0x3f3f3f3f;
11
12 struct Point {
13 double x, y;
14 Point(double a=0, double b=0): x(a), y(b) {}
15 } dot[MAX];
16
17 struct Line {
18 double up, low;
19 bool operator < (const Line& rhs) const {
20 double x1 = min(up, low);
21 double x2 = min(rhs.up, rhs.low);
22 if(x1 == x2)
23 return max(up, low) < max(rhs.up, rhs.low);
24 else
25 return x1 < x2;
26 }
27 } line[MAX];
28
29 int num[MAX], ans[MAX];
30
31 double Cross(Point A, Point B, Point C)
32 {
33 return (C.x-A.x)*(B.y-A.y)-(B.x-A.x)*(C.y-A.y);
34 }
35 int main()
36 {
37 #ifdef _Rayn
38 freopen("in.txt", "r", stdin);
39 #endif
40
41 int n, m;
42 double x1, y1, x2, y2;
43
44 while(scanf("%d", &n) != EOF && n)
45 {
46 scanf("%d%lf%lf%lf%lf", &m, &x1, &y1, &x2, &y2);
47 //printf("%d %d\n", n, m);
48 for(int i=0; i<n; ++i)
49 {
50 scanf("%lf%lf", &line[i].up, &line[i].low);
51 }
52 sort(line, line+n);
53 line[n].up = line[n].low = x2;
54
55 memset(num, 0, sizeof(num));
56 memset(ans, 0, sizeof(ans));
57 for(int i=0; i<m; ++i)
58 {
59 scanf("%lf%lf", &dot[i].x, &dot[i].y);
60 int left = 0, right = n, mid = 0;
61 while(left <= right)
62 {
63 mid = (left + right) / 2;
64 Point b(line[mid].low, y2);
65 Point c(line[mid].up, y1);
66 if(Cross(dot[i], b, c) > 0)
67 left = mid + 1;
68 else
69 right = mid - 1;
70 }
71 num[left]++;
72 }
73 for(int i=0; i<=n; ++i)
74 {
75 if(num[i])
76 ans[num[i]]++;
77 }
78 printf("Box\n");
79 for(int i=0; i<=n; ++i)
80 {
81 if(ans[i])
82 printf("%d: %d\n", i, ans[i]);
83 }
84 }
85 return 0;
86 }
POJ 2398 Toy Storage（叉积+二分）
时间： 2024-10-17 19:46:45