POJ 2900 Griddy Hobby

Griddy Hobby

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 414		Accepted: 146

Description

Attending a boring weekly session, our professor started drawing on a grid in a page of his calendar. He started at a boundary grid point P; note that P is a corner of one or two grid cells. He drew a diameter of one of those cells and continued on a straight line until reaching point Q on another edge of the grid. Then he started another line from Q, perpendicular to the line PQ until hitting another edge at point R. He kept drawing lines as above, until he could not draw a new line, either because a perpendicular line would not start with a cell diameter or it would fall on an already drawn line. Then he was puzzling how he would be able to count the number of minimal rectangles he has created on the grid. At this time, the chair of the session noticed him and asked him what he was doing. "Sorry, I was designing a problem for the ACM-ICPC Tehran site," he said.

You are to write a program to, given the dimensions R and C of a grid, the coordinates x and y of a point P on one of the edges of the grid, and the direction (up, left), (down, left), (up, right), or (down, right) of the first line, help the professor to find out the number of minimal rectangles.

Input

The input file contains multiple test cases. The first line of the input, contains t, the number of test cases that follow. Each of the following t blocks, describes a test case.

The first line of each block, contains two positive integers R and C, which are the number of horizontal and vertical gridlines, respectively (2 <= R,C <= 1000). The coordinates of the starting point comes on the next line, as two positive integers y and x, (1 <= y <= R, 1 <= x <= C ); for the upper-left point of the grid, we have x = y = 1. The third line of each test case, contains a two-character code which is one of the following:

• "DR", the first move is in the down-right direction,
• "DL", the first move is in the down-left direction,
• "UL", the first move is in the up-left direction,
• "UR", the first move is in the up-right direction.

Output

For each test case, output one line containing the number of non-overlapping rectangles formed.

Sample Input

2
4 5
2 1
DR
4 5
4 1
UR

Sample Output

1
3

Source

Tehran 2005

 

/*

Keep fighting.

计算几何题，仔细思考不难发现，所有直线交点（转折点除外，即反射点），所有当前

新生成的直线与其相邻直线外所有直线交点的个数和即为所求，所以在模拟过程中可以

存储所有的直线当处理某一条直线时，只需遍历所有已经生成的直线（其相邻直线除外）

统计交点个数即可。终止条件的判断用了个技巧，每个点都有1，2，3，4四个方向，如果

有哪个点的方向被重复或者到底矩形的四个顶点则访问终止

*/

 

#include <iostream> #include <string> #define minv(a, b) ((a) <= (b) ? (a) : (b)) #define maxv(a, b) ((a) >= (b) ? (a) : (b)) #define MAX_N 1005 using namespace std; //输入：行列宽，起始点的位置 int r, c, sr, sc, sdir; int caseN; //标记每个点四个方向的访问情况 struct grid { bool dir[4]; }grids[MAX_N + 1][MAX_N + 1]; //记录生成的直线 struct line { int r1, c1, r2, c2; }lines[MAX_N * MAX_N + 1]; int ln, countv; //ln:直线的个数，countv:矩形的数量 //表示每个点，以及从它出去的直线的方向 struct pos { int r, c, dir; }; //按照某种规则比较两个点的大小 bool compare(const pos &p1, const pos &p2) { if(p1.c < p2.c) return true; else if(p1.c == p2.c) { return p1.r <= p2.r; } else return false; } //输入是当前的以及其直线生长方向，输出是新的交点以及反射线的方向，返回值是是否继续生长，返回false表示终止 bool getNext(const pos &cur_pos, pos &new_pos) { //判断是否重复访问过当前直线 if(grids[cur_pos.r][cur_pos.c].dir[cur_pos.dir]) return false; //判断当前点是不是矩形的四个角的顶点 if(!(cur_pos.r == sr && cur_pos.c == sc) && ((cur_pos.r == 1 && cur_pos.c == 1) || (cur_pos.r == 1 && cur_pos.c == c) || (cur_pos.r == r && cur_pos.c == 1) || (cur_pos.r == r && cur_pos.c == c))) return false; //标记访问过当前直线 grids[cur_pos.r][cur_pos.c].dir[cur_pos.dir] = true; //a是当前直线的斜率，d是与y轴的交点纵坐标 int a, d; if(cur_pos.dir == 2 || cur_pos.dir == 3) a = -1; else a = 1; //计算矩形左边的交点 d = cur_pos.r - cur_pos.c * a; int newr = a * 1 + d, newc = 1; //交点不是原先的点且交点在有效范围内 if(!(newr == cur_pos.r && newc == cur_pos.c) && newr >= 1 && newr <= r) { new_pos.r = newr; new_pos.c = newc; new_pos.dir = cur_pos.dir == 3 ? 4 : 2; return true; } //计算矩形右边的交点 newr = a * c + d, newc = c; if(!(newr == cur_pos.r && newc == cur_pos.c) && newr >= 1 && newr <= r) { new_pos.r = newr; new_pos.c = newc; new_pos.dir = cur_pos.dir == 4 ? 3 : 1; return true; } //计算矩形顶边的交点 newr = 1, newc = (newr - d) / a; if(!(newr == cur_pos.r && newc == cur_pos.c) && newc >= 1 && newc <= c) { new_pos.r = newr; new_pos.c = newc; new_pos.dir = cur_pos.dir == 2 ? 4 : 3; return true; } //计算矩形底边的交点 newr = r, newc = (newr - d) / a; if(!(newr == cur_pos.r && newc == cur_pos.c) && newc >= 1 && newc <= c) { new_pos.r = newr; new_pos.c = newc; new_pos.dir = cur_pos.dir == 4 ? 2 : 1; return true; } return false; } //判断相交的代码 //差积 int cross(int x1, int y1, int x2, int y2, int x3, int y3) { return (x3 - x1) * (y2 - y1) - (x2 - x1) * (y3 - y1); } //判断点3是否在点(1,2)上 bool onLine(int x1, int y1, int x2, int y2, int x3, int y3) { return minv(x1, x2) <= x3 && x3 <= maxv(x1, x2) && minv(y1, y2) <= y3 && y3 <= maxv(y1, y2); } //判断线段[1,2]是否与线段[3,4]相交 bool intersect(int x1, int y1, int x2, int y2, int x3, int y3, int x4, int y4) { int d1 = cross(x1, y1, x2, y2, x3, y3); int d2 = cross(x1, y1, x2, y2, x4, y4); int d3 = cross(x3, y3, x4, y4, x1, y1); int d4 = cross(x3, y3, x4, y4, x2, y2); if(((d1 > 0 && d2 < 0) || (d1 < 0 && d2 > 0)) && ((d3 > 0 && d4 < 0) || (d3 < 0 && d4 > 0))) return true; if(d1 == 0 && onLine(x1, y1, x2, y2, x3, y3)) return true; if(d2 == 0 && onLine(x1, y1, x2, y2, x4, y4)) return true; if(d3 == 0 && onLine(x3, y3, x4, y4, x1, y1)) return true; if(d4 == 0 && onLine(x3, y3, x4, y4, x2, y2)) return true; return false; } //插入线段[p1,p2] void insertLine(const pos &p1, const pos &p2) { //记录新的直线 lines[ln].r1 = p1.r; lines[ln].c1 = p1.c; lines[ln].r2 = p2.r; lines[ln].c2 = p2.c; //判断与之前记录的线段的交点个数，相邻线段除外即上一个线段，注意边界 for(int i = 0; i < ln - 1; i++) { if(intersect(lines[i].r1, lines[i].c1, lines[i].r2, lines[i].c2, lines[ln].r1, lines[ln].c1, lines[ln].r2, lines[ln].c2)) countv++; //增加交点数 } ln++; } int main() { scanf("%d", &caseN); pos cur_pos, new_pos; string d; while(caseN--) { //输入部分 scanf("%d%d", &r, &c); scanf("%d%d", &sr, &sc); cin>>d; memset(grids, 0, sizeof(grids)); ln = countv = 0; //记录起始点的信息 cur_pos.r = sr; cur_pos.c = sc; if(d == "DR") cur_pos.dir = 4; else if(d == "DL") cur_pos.dir = 3; else if(d == "UL") cur_pos.dir = 1; else cur_pos.dir = 2; //边一直在走，直到不可走为止 while(getNext(cur_pos, new_pos)) { insertLine(cur_pos, new_pos); cur_pos = new_pos; } //输出 cout<<countv<<endl; } return 0; }