ACM-搜索之Red and Black——hdu1312
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Red and Black
Problem Description
There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
Input
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.
There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.
'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)
Output
For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).
Sample Input
6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0
Sample Output
45
59
6
Problem Description
There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
Input
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.
There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.
'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)
Output
For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).
Sample Input
6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0
Sample Output
45
59
6
13
很简单的一道题目,我觉得深搜广搜应该都能过= =。就是从@开始所有向四个方向走,查看。的数量,当然
走到边界返回,碰到#返回。
这道题目有一点比较坑,就是输入的两个数,第一个代表的是列数第二个是行数,我就因为这个纠结了会,
后来调试才发现这个问题,o(╯□╰)o囧啊。。。。
DFS:
#include <iostream>#include <string.h>using namespace std;// mapp存储地图,dis数组存储四个方向,vis存储是否走过char mapp[21][21];int n,m,num,vis[21][21],dis[4][2]={-1,0,0,1,0,-1,1,0};// 判断是否越界,或者是否碰到#,或者是否已经遍历过bool judge(int x,int y){ if(x<0 || y<0 ||x>=m || y>=n) return 0; if(mapp[x][y]=='#' || vis[x][y]==1) return 0; return 1;}void dfs(int x,int y){ int i,xx,yy; for(i=0;i<4;++i) { xx=x+dis[i][0]; yy=y+dis[i][1]; if(judge(xx,yy)) { vis[xx][yy]=1; num+=1; dfs(xx,yy); } }}int main(){ int i,j,s_x,s_y; while(cin>>n>>m) { if(n==0 && m==0) break; // 输入地图,记录起始点 for(i=0;i<m;++i) for(j=0;j<n;++j) { cin>>mapp[i][j]; if(mapp[i][j]=='@'){s_x=i;s_y=j;} } memset(vis,0,sizeof(vis)); // 从第一个开始走,第一步要算上 vis[s_x][s_y]=1; num=1; dfs(s_x,s_y); cout<<num<<endl; } return 0;}
BFS:
#include <iostream>using namespace std;// mapp存储地图,dis数组存储四个方向,vis存储是否走过char mapp[21][21];int n,m;int bfs(int x,int y){ if(x<0 || y<0 ||x>=m || y>=n) return 0; if(mapp[x][y]=='#') return 0; mapp[x][y]='#'; return bfs(x-1,y)+bfs(x+1,y)+bfs(x,y-1)+bfs(x,y+1)+1;}int main(){ int i,j,s_x,s_y; while(cin>>n>>m) { if(n==0 && m==0) break; // 输入地图,记录起始点 for(i=0;i<m;++i) for(j=0;j<n;++j) { cin>>mapp[i][j]; if(mapp[i][j]=='@'){s_x=i;s_y=j;} } int num=bfs(s_x,s_y); cout<<num<<endl; } return 0;}
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