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.

計算幾何題，仔細思考不難發現，所有直線交點（轉折點除外，即反射點），所有當前

新生成的直線與其相鄰直線外所有直線交點的個數和即爲所求，所以在模擬過程中可以

存儲所有的直線當處理某一條直線時，只需遍歷所有已經生成的直線（其相鄰直線除外）

統計交點個數即可。終止條件的判斷用了個技巧，每個點都有1，2，3，4四個方向，如果

有哪個點的方向被重複或者到底矩形的四個頂點則訪問終止

*/

 

#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; //輸入：行列寬，起始點的位置 int r, c, sr, sc, sdir; int caseN; //標記每個點四個方向的訪問情況 struct grid { bool dir[4]; }grids[MAX_N + 1][MAX_N + 1]; //記錄生成的直線 struct line { int r1, c1, r2, c2; }lines[MAX_N * MAX_N + 1]; int ln, countv; //ln:直線的個數，countv:矩形的數量 //表示每個點，以及從它出去的直線的方向 struct pos { int r, c, dir; }; //按照某種規則比較兩個點的大小 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; } //輸入是當前的以及其直線生長方向，輸出是新的交點以及反射線的方向，返回值是是否繼續生長，返回false表示終止 bool getNext(const pos &cur_pos, pos &new_pos) { //判斷是否重複訪問過當前直線 if(grids[cur_pos.r][cur_pos.c].dir[cur_pos.dir]) return false; //判斷當前點是不是矩形的四個角的頂點 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; //標記訪問過當前直線 grids[cur_pos.r][cur_pos.c].dir[cur_pos.dir] = true; //a是當前直線的斜率，d是與y軸的交點縱座標 int a, d; if(cur_pos.dir == 2 || cur_pos.dir == 3) a = -1; else a = 1; //計算矩形左邊的交點 d = cur_pos.r - cur_pos.c * a; int newr = a * 1 + d, newc = 1; //交點不是原先的點且交點在有效範圍內 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; } //計算矩形右邊的交點 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; } //計算矩形頂邊的交點 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; } //計算矩形底邊的交點 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; } //判斷相交的代碼 //差積 int cross(int x1, int y1, int x2, int y2, int x3, int y3) { return (x3 - x1) * (y2 - y1) - (x2 - x1) * (y3 - y1); } //判斷點3是否在點(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); } //判斷線段[1,2]是否與線段[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; } //插入線段[p1,p2] void insertLine(const pos &p1, const pos &p2) { //記錄新的直線 lines[ln].r1 = p1.r; lines[ln].c1 = p1.c; lines[ln].r2 = p2.r; lines[ln].c2 = p2.c; //判斷與之前記錄的線段的交點個數，相鄰線段除外即上一個線段，注意邊界 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++; //增加交點數 } ln++; } int main() { scanf("%d", &caseN); pos cur_pos, new_pos; string d; while(caseN--) { //輸入部分 scanf("%d%d", &r, &c); scanf("%d%d", &sr, &sc); cin>>d; memset(grids, 0, sizeof(grids)); ln = countv = 0; //記錄起始點的信息 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; //邊一直在走，直到不可走爲止 while(getNext(cur_pos, new_pos)) { insertLine(cur_pos, new_pos); cur_pos = new_pos; } //輸出 cout<<countv<<endl; } return 0; }