原题链接 Red and Black Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 33776 Accepted: 18319 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) The end of the input is indicated by a line consisting of two zeros. 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 Source
Japan 2004 Domestic 题意:有一个矩形的房间,里面铺满了方形的瓷砖。每一个瓷砖有红色还有黑色两种,人只能站在黑色的瓷砖上不能站在红色的瓷砖上,每次只能走前后左右四个方向的四个瓷砖,计算能够到达的黑色瓷砖的总面积数。 思路:深度优先搜索,走过的地方直接变成红色
#include <cstdio>#include <cstring>#include <iostream>using namespace std;int dx[4]={1,0,-1,0};int dy[4]={0,-1,0,1};int n,m;char a[22][22];bool used[22][22];void dfs(int &res,int x,int y){ used[x][y]=true; res++; for(int i=0;i<4;i++){ int nx=x+dx[i],ny=y+dy[i]; if(nx >= 0 && nx <n && ny >= 0 && ny < m && !used[nx][ny]){ dfs(res,nx,ny); //心得:对于这种染色而不是寻找路径的这种问题,我们就不需要每次令used[x][y]=false; } } return;}int main(){ while(scanf("%d%d",&m,&n)==2 && n+m!=0){ int x,y; memset(used,false,sizeof(used)); for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ cin >> a[i][j]; if(a[i][j]=='@'){ x=i;y=j; } else if(a[i][j]=='#'){ used[i][j]=true; } } } int res=0; dfs(res,x,y); cout << res << endl; } return 0;}新闻热点
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