poj_1111Image Perimeters
来源:互联网 发布:天翼云直链平台 源码 编辑:程序博客网 时间:2024/05/23 01:24
Image Perimeters
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 6891 Accepted: 4106
Description
Technicians in a pathology lab analyze digitized images of slides. Objects on a slide are selected for analysis by a mouse click on the object. The perimeter of the boundary of an object is one useful measure. Your task is to determine this perimeter for selected objects.
The digitized slides will be represented by a rectangular grid of periods, '.', indicating empty space, and the capital letter 'X', indicating part of an object. Simple examples are
An X in a grid square indicates that the entire grid square, including its boundaries, lies in some object. The X in the center of the grid below is adjacent to the X in any of the 8 positions around it. The grid squares for any two adjacent X's overlap on an edge or corner, so they are connected.
An object consists of the grid squares of all X's that can be linked to one another through a sequence of adjacent X's. In Grid 1, the whole grid is filled by one object. In Grid 2 there are two objects. One object contains only the lower left grid square. The remaining X's belong to the other object.
The technician will always click on an X, selecting the object containing that X. The coordinates of the click are recorded. Rows and columns are numbered starting from 1 in the upper left hand corner. The technician could select the object in Grid 1 by clicking on row 2 and column 2. The larger object in Grid 2 could be selected by clicking on row 2, column 3. The click could not be on row 4, column 3.
One useful statistic is the perimeter of the object. Assume each X corresponds to a square one unit on each side. Hence the object in Grid 1 has perimeter 8 (2 on each of four sides). The perimeter for the larger object in Grid 2 is illustrated in the figure at the left. The length is 18.
Objects will not contain any totally enclosed holes, so the leftmost grid patterns shown below could NOT appear. The variations on the right could appear:
The digitized slides will be represented by a rectangular grid of periods, '.', indicating empty space, and the capital letter 'X', indicating part of an object. Simple examples are
XX Grid 1 .XXX Grid 2 XX .XXX .XXX ...X ..X. X...
An X in a grid square indicates that the entire grid square, including its boundaries, lies in some object. The X in the center of the grid below is adjacent to the X in any of the 8 positions around it. The grid squares for any two adjacent X's overlap on an edge or corner, so they are connected.
XXX XXX Central X and adjacent X's XXX
An object consists of the grid squares of all X's that can be linked to one another through a sequence of adjacent X's. In Grid 1, the whole grid is filled by one object. In Grid 2 there are two objects. One object contains only the lower left grid square. The remaining X's belong to the other object.
The technician will always click on an X, selecting the object containing that X. The coordinates of the click are recorded. Rows and columns are numbered starting from 1 in the upper left hand corner. The technician could select the object in Grid 1 by clicking on row 2 and column 2. The larger object in Grid 2 could be selected by clicking on row 2, column 3. The click could not be on row 4, column 3.
One useful statistic is the perimeter of the object. Assume each X corresponds to a square one unit on each side. Hence the object in Grid 1 has perimeter 8 (2 on each of four sides). The perimeter for the larger object in Grid 2 is illustrated in the figure at the left. The length is 18.
Objects will not contain any totally enclosed holes, so the leftmost grid patterns shown below could NOT appear. The variations on the right could appear:
Impossible Possible XXXX XXXX XXXX XXXX X..X XXXX X... X... XX.X XXXX XX.X XX.X XXXX XXXX XXXX XX.X ..... ..... ..... ..... ..X.. ..X.. ..X.. ..X.. .X.X. .XXX. .X... ..... ..X.. ..X.. ..X.. ..X.. ..... ..... ..... .....
Input
The input will contain one or more grids. Each grid is preceded by a line containing the number of rows and columns in the grid and the row and column of the mouse click. All numbers are in the range 1-20. The rows of the grid follow, starting on the next line, consisting of '.' and 'X' characters.
The end of the input is indicated by a line containing four zeros. The numbers on any one line are separated by blanks. The grid rows contain no blanks.
The end of the input is indicated by a line containing four zeros. The numbers on any one line are separated by blanks. The grid rows contain no blanks.
Output
For each grid in the input, the output contains a single line with the perimeter of the specified object.
Sample Input
2 2 2 2XXXX6 4 2 3.XXX.XXX.XXX...X..X.X...5 6 1 3.XXXX.X....X..XX.X.X...X..XXX.7 7 2 6XXXXXXXXX...XXX..X..XX..X...X..X..XX.....XXXXXXXX7 7 4 4XXXXXXXXX...XXX..X..XX..X...X..X..XX.....XXXXXXXX0 0 0 0
Sample Output
81840488
#include <iostream>#include <cstring>using namespace std;#pragma warning(disable : 4996) #define MAX 25char map[MAX][MAX];int visit[MAX][MAX];int dir[8][2]={{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1},{-1,0},{-1,1}};int r, c, sum;void dfs(int x, int y){visit[x][y] = 1;for(int i = 0; i < 8; i++){int p = x + dir[i][0];int q = y + dir[i][1];if(p >= 1 && p <= r && q >= 1 && q <= c){if(map[p][q] == 'X' && visit[p][q] == 0){dfs(p, q);}else if(map[p][q]=='.' && (p == x || q == y)){sum++;}}else if (p == x || q == y) {sum++;}}}int main(){freopen("in.txt", "r", stdin);int i, j, x, y;while(cin >> r >> c >> x >> y){if(x + y + r + c == 0) break;for(i = 1; i <= r; i++){for(j = 1; j <= c; j++){visit[i][j] = 0;}}for(i=1; i <= r; i++){for(j = 1; j <= c; j++){cin >> map[i][j];}}sum = 0;dfs(x, y);cout << sum << endl;}return 0;}
- poj_1111Image Perimeters
- poj1111Image Perimeters
- Image Perimeters
- pku_1111_Image Perimeters
- Image Perimeters
- Image Perimeters
- ZOJ 1047 Image Perimeters
- POJ 1111 Image Perimeters
- poj 1111 Image Perimeters
- 1111 Image Perimeters
- ZOJ1047 Image Perimeters
- POJ 1111 Image Perimeters
- Poj 1111 Image Perimeters
- POJ 1111 - Image Perimeters
- poj 1111 Image Perimeters
- POJ1111:Image Perimeters(DFS)
- POJ-1111-Image Perimeters
- poj1111 Image Perimeters
- 重拾ssh之hibernate配置
- HDU 1873 看病要排队 优先队列
- Android Gallery获取滑动停止的位置
- Linux下内存释放问题
- WebSocket 的 PHP 实现 - phpwebsocket
- poj_1111Image Perimeters
- 主函数 main WinMain _tmain _tWinMain 的区别
- Fiddler (数据抓包)跟踪监控android数据包工具使用
- WebKit Layout (布局)
- Ubuntu 12.10升级
- call
- error C2664: MessageBoxW不能将参数2const char*转换为LPCWSTR的解决办法
- a utility to compare date diff and date advance^_^
- 获取本地IP