POJ 1111 Image Perimeters【单身快乐】【Bfs】
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Image Perimeters
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 8684 Accepted: 5197
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
题目大意:给你一个n*m的图,和一个起点,让你求一下这个起点所包含的一个以X组成的形状的周长。
思路:因为图比较小,直接Bfs,找到起点所连通的联通块,对于vis【】【】数组,其1表示是形状的一部分,0表示不是形状的一部分。那么题目所求就是这个联通块所组成的形状的周长,对于如何求周长这个处理,其实小伙伴们认真思考一下就不难得到结论,求一下所有vis【】【】==1的部分的四个方向(上,下,左,右)是否有0,如果有0,那么这个方向上就有一条线段是这个形状的周长之一。那么我们只要求一下这个图的vis【】【】==1的四个方向0的个数和即可。
单身题,愿各位单身狗少撸题,多撩妹。在新的一天有完美的偶遇。
AC代码:
#include<stdio.h>#include<string.h>#include<queue>using namespace std;struct zuobiao{ int x,y;}now,nex;char a[100][100];int vis[100][100];int fx[8]={0,0,1,-1,1,-1,1,-1};int fy[8]={1,-1,0,0,1,-1,-1,1};int n,m,sx,sy;void Bfs(){ queue<zuobiao >s; memset(vis,0,sizeof(vis)); now.x=sx; now.y=sy; s.push(now); vis[sx][sy]=1; while(!s.empty()) { now=s.front(); s.pop(); for(int i=0;i<8;i++) { nex.x=now.x+fx[i]; nex.y=now.y+fy[i]; if(nex.x>=1&&nex.x<=n&&nex.y>=1&&nex.y<=m&&vis[nex.x][nex.y]==0&&a[nex.x][nex.y]=='X') { vis[nex.x][nex.y]=1; s.push(nex); } } }}int main(){ while(~scanf("%d%d%d%d",&n,&m,&sx,&sy)) { if(n+m+sx+sy==0)break; for(int i=1;i<=n;i++) { scanf("%s",a[i]+1); } Bfs(); int output=0; for(int i=0;i<=n+1;i++) { for(int j=0;j<=m+1;j++) { if(vis[i][j]==1) { for(int k=0;k<4;k++) { int xx=i+fx[k]; int yy=j+fy[k]; if(vis[xx][yy]==0)output++; } } } } printf("%d\n",output); }}
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