Toy Storage
Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box to put his toys in. Unfortunately, Reza 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 Reza to find his favorite toys anymore.
Reza's parents came up with the following idea. They put cardboard partitions into the box. Even if Reza keeps throwing his toys into the box, at least toys that get thrown into different partitions stay separate. The box looks like this from the top:
We want for each positive integer t, such that there exists a partition with t toys, determine how many partitions have t, toys.
Input
The input consists of a number of cases. The first line consists of six integers n, m, x1, y1, x2, y2. The number of cardboards to form the partitions is n (0 < n <= 1000) and the number of toys is given in m (0 < m <= 1000). 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 each consists of two integers Ui Li, indicating that the ends of the ith cardboard is at the coordinates (Ui, y1) and (Li, y2). You may assume that the cardboards do not intersect with each other. The next m lines each consists of two integers Xi Yi specifying where the ith toy has landed in the box. You may assume that no toy will land on a cardboard.
A line consisting of a single 0 terminates the input.
Output
For each box, first provide a header stating "Box" on a line of its own. After that, there will be one line of output per count (t > 0) of toys in a partition. The value t will be followed by a colon and a space, followed the number of partitions containing t toys. Output will be sorted in ascending order of t for each box.
Sample Input
4 10 0 10 100 0
20 20
80 80
60 60
40 40
5 10
15 10
95 10
25 10
65 10
75 10
35 10
45 10
55 10
85 10
5 6 0 10 60 0
4 3
15 30
3 1
6 8
10 10
2 1
2 8
1 5
5 5
40 10
7 9
0
Sample Output
Box
2: 5
Box
1: 4
2: 1
Source
题目类型：叉积+二分枚举
算法分析：使用叉积来判断点在一条直线的左右侧，可以二分枚举直线来判断。注意在枚举前要先将直线按照从左至右的顺序排好序！！！
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 |
/************************************************* Author :supermaker Created Time :2016/1/24 14:00:45 File Location :C:\Users\abcd\Desktop\TheEternalPoet **************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1000 + 66; struct Point { double x, y; Point () {} Point (double xx, double yy) : x (xx), y (yy) {} Point operator - (const Point &a) const { return Point (x - a.x, y - a.y); } double operator ^ (const Point &a) const { return x * a.y - y * a.x; } double operator * (const Point &a) const { return x * a.x + y * a.y; } }; struct Line { Point s, e; Line () {} Line (Point ss, Point ee) : s (ss), e (ee) {} }; Line line[maxn]; int res[maxn], cnt[maxn], n, m; double xmult (Point p1, Point p2, Point p3) { return (p2 - p1) ^ (p3 - p1); } int sgn (double val) { if (fabs (val) < EPS) return 0; else if (val < 0) return -1; return 1; } bool cmp (Line a, Line b) { return a.s.x < b.s.x; } int main() { //CFF; //CPPFF; while (scanf ("%d", &n) != EOF) { if (n == 0) break; scanf ("%d", &m); double x1, y1, x2, y2; scanf ("%lf%lf%lf%lf", &x1, &y1, &x2, &y2); for (int i = 1; i <= n; i++) { double Ui, Li; scanf ("%lf%lf", &Ui, &Li); line[i] = Line (Point (Ui, y1), Point (Li, y2)); } sort (line + 1, line + 1 + n, cmp); memset (res, 0, sizeof (res)); memset (cnt, 0, sizeof (cnt)); for (int i = 1; i <= m; i++) { double x, y; scanf ("%lf%lf", &x, &y); int low = 1, high = n; while (low <= high) { int mid = (low + high) >> 1; double tt = xmult (line[mid].s, Point (x, y), line[mid].e); if (sgn (tt) == 1) high = mid - 1; else low = mid + 1; } res[max(low,high)]++; } puts ("Box"); int maxval = -INF; for (int i = 1; i <= n + 1; i++) { cnt[res[i]]++; if (res[i]) maxval = max (maxval, res[i]); } for (int i = 1; i <= maxval; i++) if (cnt[i]) printf ("%d: %d\n", i, cnt[i]); } return 0; } |
- « 上一篇：poj2318
- poj1269：下一篇 »