POJ-1111---Image Perimeters (bfs)
来源:互联网 发布:淘宝风力灭火机价格 编辑:程序博客网 时间:2024/06/11 07:48
Image Perimeters
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 9255 Accepted: 5458
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
Source
Mid-Central USA 2001
题意:有一个n*m的矩阵,给一个起点(x,y),求这个点所在连通块的周长,算周长的方法题中图很清楚;
思路:bfs遍历连通块中的每一个点,对于每个点,看它的上下左右四个方向,不为‘X'则周长+1,具体看代码;
AC代码:
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<queue>#define inf 0xfffffffusing namespace std;int n,m;int x,y;int ans,cnt;struct node{ int x; int y;}p,q,tmp;queue<node> que;int next[8][2]={-1,-1,-1,0,-1,1,0,1,0,-1,1,-1,1,0,1,1};int _next[4][2]={1,0,-1,0,0,1,0,-1};char a[25][25];int vis[25][25], _vis[25][25];bool check(int x,int y){ if(x<1||x>n||y<1||y>m)return false; if(vis[x][y]==1)return false; if(a[x][y]!='X')return false; return true;}void bfs(){ while(que.size()) { tmp=que.front(); que.pop(); for(int i=0;i<4;i++)//遍历连通块每个点的四个方向 { int xx=tmp.x+_next[i][0]; int yy=tmp.y+_next[i][1]; if(xx<1||xx>n||yy<1||yy>m||a[xx][yy]!='X')//符合条件周长+1 ans++; } for(int i=0;i<=7;i++)//遍历当前点的8个方向,若有覆盖和条件的点就入队 { int tx=tmp.x+next[i][0]; int ty=tmp.y+next[i][1]; if(check(tx,ty)) { q.x=tx; q.y=ty; que.push(q); vis[tx][ty]=1; } } }}int main(){ while(~scanf("%d%d",&n,&m)) { scanf("%d%d",&x,&y); if(n==0&&m==0&&x==0&&y==0)break; for(int i=1;i<=n;i++) scanf("%s",a[i]+1); memset(vis,0,sizeof(vis)); memset(_vis,0,sizeof(_vis)); if(a[x][y]=='X')//判断起点,若为'.',则直接输出-1; { p.x=x; p.y=y; que.push(p);//起点入队,并标记 vis[x][y]=1; ans=0; bfs(); printf("%d\n",ans); } else { printf("0\n"); } } return 0;}
阅读全文
0 0
- POJ-1111---Image Perimeters (bfs)
- POJ 1111 Image Perimeters【单身快乐】【Bfs】
- POJ 1111 Image Perimeters
- poj 1111 Image Perimeters
- POJ 1111 Image Perimeters
- Poj 1111 Image Perimeters
- POJ 1111 - Image Perimeters
- poj 1111 Image Perimeters
- POJ-1111-Image Perimeters
- poj 1111 Image Perimeters
- POJ 1111 Image Perimeters
- POJ:1111 Image Perimeters
- poj 1111:Image Perimeters
- POJ 1111 Image Perimeters
- POJ Image Perimeters 1111
- poj 1111 Image Perimeters
- POJ - 1111 Image Perimeters
- POJ 1111 Image Perimeters
- Tomcat/Jsp,解决'怎么访问本地图片/资源?'
- xml2json 转换
- Spring-Data-Jpa
- select学习及循环服务器实现
- ActivityStackSupervisor.java关键信息解读
- POJ-1111---Image Perimeters (bfs)
- Day7-19.Coding style
- P1049 装箱问题
- Day7-20.Using Java operators
- linux查看文件内容
- kali 漏洞利用metasploit
- Day7-21.Precedence
- HDU 6127 Hard challenge(几何 多校第七场)
- Java-栈方法实现迷宫问题