Time Limit: 2000MS
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.
题意：
一个矩形被n条线段分成了n+1个区域，然后询问每个区域里有几个玩具。
直接对玩具最左端的线段进行二分。
#include "stdio.h" #include "iostream" #include "algorithm" #include "math.h" #include "string.h" using namespace std; const int maxn = 1e5 + 10; struct Point { int x, y; Point(int x = 0, int y = 0):x(x), y(y) {} void in(void) { scanf("%d%d", &x, &y);} }; struct Line { Point a, b; int ang; Line() {} Line(const Line& L): a(L.a), b(L.b), ang(L.ang) {} Line(const Point& a,const Point& b):a(a),b(b){ if (a.x == b.x) ang = 0; else if ((a.y - b.y)/(a.x - b.y) < 0) ang = -1; else ang = 1; } bool operator < (const Line& L) const { return a.x<L.a.x; } }; Line line[maxn]; int res[maxn]; int n, m; bool JudgePoint(Line L1, Point c, int mid) { int flag = (L1.a.x - L1.b.x)*(L1.a.y - L1.b.y); if (flag == 0) return L1.a.x <= c.x; if (L1.a.x - L1.b.x > 0){ if (c.y*(L1.a.x - L1.b.x) > (L1.a.y - L1.b.y)*(c.x - L1.b.x)+L1.b.y*(L1.a.x - L1.b.x)) return flag < 0; else return flag > 0; } else { if (c.y*(L1.a.x - L1.b.x) < (L1.a.y - L1.b.y)*(c.x - L1.b.x)+L1.b.y*(L1.a.x - L1.b.x)) return flag < 0; else return flag > 0; } } void solve(Point c) { int ub = n + 1, lb = 0; int t; while (ub >= lb) { int mid = (ub + lb) >> 1; if (JudgePoint(line[mid], c, mid)) { t = mid; lb = mid + 1; } else ub = mid - 1; } res[t]++; } int main(int argc, char const *argv[]) { int x1, y1, x0, y0, xu, xd; while (scanf("%d", &n), n) { scanf("%d", &m); memset(res, 0, sizeof(res)); scanf("%d%d%d%d", &x0, &y1, &x1, &y0); line[0] = Line(Point(x0, y0), Point(x0, y1)); line[n + 1] = Line(Point(x1, y0), Point(x1, y1)); for (int i = 1; i <= n; i++) { scanf("%d%d", &xu, &xd); line[i] = Line(Point(xd, y0), Point(xu, y1)); } sort(line, line+(n+1)); Point p; for (int i = 0; i < m; i++) { p.in(); solve(p); } for (int i = 0; i <= n; i++) { printf("%d: %d ", i, res[i]); } printf(" "); } return 0; }
和上道题一样，只是这个做了一个统计。
#include "stdio.h" #include "iostream" #include "algorithm" #include "math.h" #include "string.h" using namespace std; const int maxn = 1e5 + 10; int num[maxn]; struct Point { int x, y; Point(int x = 0, int y = 0):x(x), y(y) {} void in(void) { scanf("%d%d", &x, &y);} }; struct Line { Point a, b; int ang; Line() {} Line(const Line& L): a(L.a), b(L.b), ang(L.ang) {} Line(const Point& a,const Point& b):a(a),b(b){ if (a.x == b.x) ang = 0; else if ((a.y - b.y)/(a.x - b.y) < 0) ang = -1; else ang = 1; } bool operator < (const Line& L) const { return a.x<L.a.x; } }; Line line[maxn]; int res[maxn]; int n, m; bool JudgePoint(Line L1, Point c, int mid) { int flag = (L1.a.x - L1.b.x)*(L1.a.y - L1.b.y); if (flag == 0) return L1.a.x <= c.x; if (L1.a.x - L1.b.x > 0){ if (c.y*(L1.a.x - L1.b.x) > (L1.a.y - L1.b.y)*(c.x - L1.b.x)+L1.b.y*(L1.a.x - L1.b.x)) return flag < 0; else return flag > 0; } else { if (c.y*(L1.a.x - L1.b.x) < (L1.a.y - L1.b.y)*(c.x - L1.b.x)+L1.b.y*(L1.a.x - L1.b.x)) return flag < 0; else return flag > 0; } } void solve(Point c) { int ub = n + 1, lb = 0; int t; while (ub >= lb) { int mid = (ub + lb) >> 1; if (JudgePoint(line[mid], c, mid)) { t = mid; lb = mid + 1; } else ub = mid - 1; } res[t]++; } int main(int argc, char const *argv[]) { int x1, y1, x0, y0, xu, xd; while (scanf("%d", &n), n) { scanf("%d", &m); memset(res, 0, sizeof(res)); scanf("%d%d%d%d", &x0, &y1, &x1, &y0); line[0] = Line(Point(x0, y0), Point(x0, y1)); line[n + 1] = Line(Point(x1, y0), Point(x1, y1)); for (int i = 1; i <= n; i++) { scanf("%d%d", &xu, &xd); line[i] = Line(Point(xd, y0), Point(xu, y1)); } sort(line, line+(n+1)); Point p; for (int i = 0; i < m; i++) { p.in(); solve(p); } memset(num, 0, sizeof(num)); printf("Box "); for (int i = 0; i <= n; i++) num[res[i]]++; for (int i = 1; i <= n; i++) { if (num[i] != 0) printf("%d: %d ", i, num[i]); } } return 0; }