hdu 1047 Image Perimeters
2016-07-12 11:33
260 查看
Image Perimeters
Status
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.
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:
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.
For each grid in the input, the output contains a single line with the perimeter of the specified object.
Example input:
2 2 2 2
XX
XX
6 4 2 3
.XXX
.XXX
.XXX
...X
..X.
X...
5 6 1 3
.XXXX.
X....X
..XX.X
.X...X
..XXX.
7 7 2 6
XXXXXXX
XX...XX
X..X..X
X..X...
X..X..X
X.....X
XXXXXXX
7 7 4 4
XXXXXXX
XX...XX
X..X..X
X..X...
X..X..X
X.....X
XXXXXXX
0 0 0 0
Example output:
8
18
40
48
8
Source
Mid-Central USA 2001
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <math.h>
using namespace std;
#define maxn 150
int n,m,xx,yy;
int dx[]={0,0,1,-1,-1,1,-1,1};
int dy[]={1,-1,0,0,-1,1,1,-1};
char s[maxn][maxn];
int vis[maxn][maxn];
int ans;
bool check(int x,int y)
{
if(x>=n || x<0 ||y>=m || y<0)
return 0;
return 1;
}
void dfs(int x,int y)
{
ans+=4;
for(int i=0;i<4;i++)
{
int xx=x+dx[i];
int yy=y+dy[i];
if(check(xx,yy) && s[xx][yy]=='X')
ans--;
}
vis[x][y]=1;
for(int i=0;i<8;i++)
{
int xx=x+dx[i];
int yy=y+dy[i];
if(check(xx,yy) && s[xx][yy]=='X' && !vis[xx][yy])
{
dfs(xx,yy);
}
}
}
int main()
{
while(~scanf("%d%d%d%d",&n,&m,&xx,&yy))
{
if(n==0 && m==0 && xx==0 && yy==0)
break;
ans=0;
memset(vis,0,sizeof(vis));
for(int i=0;i<n;i++)
scanf("%s",s[i]);
dfs(xx-1,yy-1);
cout<<ans<<endl;
}
return 0;
}
Time Limit: 2MS | Memory Limit: 65536KB | 64bit IO Format: %lld & %llu |
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.
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:
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.
For each grid in the input, the output contains a single line with the perimeter of the specified object.
Example input:
2 2 2 2
XX
XX
6 4 2 3
.XXX
.XXX
.XXX
...X
..X.
X...
5 6 1 3
.XXXX.
X....X
..XX.X
.X...X
..XXX.
7 7 2 6
XXXXXXX
XX...XX
X..X..X
X..X...
X..X..X
X.....X
XXXXXXX
7 7 4 4
XXXXXXX
XX...XX
X..X..X
X..X...
X..X..X
X.....X
XXXXXXX
0 0 0 0
Example output:
8
18
40
48
8
Source
Mid-Central USA 2001
#include <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <math.h>
using namespace std;
#define maxn 150
int n,m,xx,yy;
int dx[]={0,0,1,-1,-1,1,-1,1};
int dy[]={1,-1,0,0,-1,1,1,-1};
char s[maxn][maxn];
int vis[maxn][maxn];
int ans;
bool check(int x,int y)
{
if(x>=n || x<0 ||y>=m || y<0)
return 0;
return 1;
}
void dfs(int x,int y)
{
ans+=4;
for(int i=0;i<4;i++)
{
int xx=x+dx[i];
int yy=y+dy[i];
if(check(xx,yy) && s[xx][yy]=='X')
ans--;
}
vis[x][y]=1;
for(int i=0;i<8;i++)
{
int xx=x+dx[i];
int yy=y+dy[i];
if(check(xx,yy) && s[xx][yy]=='X' && !vis[xx][yy])
{
dfs(xx,yy);
}
}
}
int main()
{
while(~scanf("%d%d%d%d",&n,&m,&xx,&yy))
{
if(n==0 && m==0 && xx==0 && yy==0)
break;
ans=0;
memset(vis,0,sizeof(vis));
for(int i=0;i<n;i++)
scanf("%s",s[i]);
dfs(xx-1,yy-1);
cout<<ans<<endl;
}
return 0;
}
相关文章推荐
- Win2003利用dfs(分布式文件系统)在负载均衡下的文件同步配置方案
- win2003分布式文件系统(dfs)配置方法[图文详解]
- win2003分布式文件系统及其部署 图文教程
- Hadoop2.6+jdk8的安装部署(1)——使用jar包安装部署【详细】
- Hadoop FS Shell
- DFS使用方法总结
- FastDFS注意事项
- 无忧技术带您预览DFS(分布式文件系统)管理控制台
- C 语言实现迷宫 DFS算法
- 一幅图弄清DFT与DTFT,DFS的关系
- HDFS---Namenode
- HDFS ---- Services startup
- POJ1523 SPF dfs
- poj1731 Orders dfs
- Surrounded Regions
- Binary Tree Zigzag Level Order Traversal,Restore IP Addresses,Word Search,Simplify Path
- DFS基础(1)
- HDU1241 Oil Deposits
- DFS算法有趣小题目
- HDU 1016 Prime Ring Problem