poj1979
来源:互联网 发布:c类超高速usb 3.0端口 编辑:程序博客网 时间:2024/05/20 20:22
Red and Black
Time Limit: 1000MS
Memory Limit: 30000K
Total Submissions: 22633
Accepted: 12217
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
4559613
题意:
有@出发能够扫到的所有的‘.’(不能通过“#”)
搜堆栈题搜到的,但是比较像搜索,就先用dfs写了一遍,后来看也有用bfs、递归和堆栈写的,递归的跟dfs很像,堆栈的方法也就是把bfs里的队列换成了堆栈,思想都一样,只写了递归的,堆栈的就没写。
首先是dfs:
Source Code
Problem: 1979 User: Memory: 212K Time: 0MSLanguage: C++ Result: Accepted- Source Code
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cstdlib>using namespace std;const int maxn=100;int row,line;int table[maxn][maxn],visit[maxn][maxn],c;int move[4][4]={{0,1},{0,-1},{1,0},{-1,0}};void dfs(int l,int r){ for(int i=0;i<4;i++) { //cout<<i<<endl; int tmpl=l+move[i][0],tmpr=r+move[i][1]; if(tmpl<=line && tmpl>0 && tmpr>0 &&tmpr <=row && table[tmpl][tmpr] && !visit[tmpl][tmpr]){ visit[tmpl][tmpr]=1; c++; dfs(tmpl,tmpr); //cout<<tmpl<<" "<<tmpr<<endl; } } return;}int main(){ char tmp,test; int al,ar; while(scanf("%d %d",&row,&line)&&row&&line) { al=1;ar=1; fill(visit[0],visit[maxn],0); fill(table[0],table[maxn],0); c=1; for(int i=1;i<=line;i++) { getchar(); for(int j=1;j<=row;j++) { scanf("%c",&tmp); if(tmp=='@') { al=i;ar=j;visit[i][j]=1;table[i][j]=2; } if(tmp=='.') table[i][j]=1; } } dfs(al,ar); printf("%d\n",c); }}
然后是bfs,稍微烦了点,用G++提交才过的,C++会RE,不知道为什么
Source Code
Problem: 1979
User:
Memory: 672K
Time: 16MS
Language: G++
Result: Accepted
· Source Code
#include <iostream>#include <cstdio>#include <cstring>using namespace std;const int maxn=30;int row,line,front=-1,rear=-1;int table[maxn][maxn],visit[maxn][maxn],c;int move[4][4]={{0,1},{0,-1},{1,0},{-1,0}};struct queue{ int l; int r;};queue q[1000];void bfs(){ int tmpl,tmpr,templ,tempr; while(front!=rear) { rear++; templ=q[rear].l; tempr=q[rear].r; for(int i=0;i<4;i++) { tmpl=templ+move[i][0],tmpr=tempr+move[i][1]; if(tmpl<=line && tmpl>0 && tmpr>0 &&tmpr <=row && table[tmpl][tmpr] && !visit[tmpl][tmpr]){ visit[tmpl][tmpr]=1; q[++front].l=tmpl; q[front].r=tmpr; c++; } } }}int main(){ int al,ar; char tmp; while(scanf("%d %d",&row,&line)&&row) { al=1;ar=1; fill(visit[0],visit[maxn],0); fill(table[0],table[maxn],0); c=1; for(int i=1;i<=line;i++) { getchar(); for(int j=1;j<=row;j++) { scanf("%c",&tmp); if(tmp=='@') { al=i;ar=j; visit[al][ar]=1; table[al][ar]=2; } if(tmp=='.') table[i][j]=1; } } q[++front].l=al; q[front].r=ar; bfs(); printf("%d\n",c); }}
递归的方法,比较简洁,但是思想和dfs差不多
Source Code
Problem: 1979
User:
Memory: 724K
Time: 0MS
Language: G++
Result: Accepted
· Source Code
#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cstdlib>using namespace std;const int maxn=100;int row,line;int table[maxn][maxn],c;int count(int l,int r){ if(l<=0 || l>line || r<=0 || r>row || !table[l][r]) return 0; table[l][r]=0; return 1+count(l+1,r)+count(l-1,r)+count(l,r+1)+count(l,r-1);}int main(){ char tmp; int al,ar; while(scanf("%d %d",&row,&line)&&row&&line) { al=1;ar=1; fill(table[0],table[maxn],0); c=1; for(int i=1;i<=line;i++) { getchar(); for(int j=1;j<=row;j++) { scanf("%c",&tmp); if(tmp=='@') { al=i;ar=j;table[i][j]=2; } if(tmp=='.') table[i][j]=1; } } printf("%d\n",count(al,ar)); }}
- POJ1979
- poj1979
- POJ1979
- POJ1979
- poj1979
- poj1979
- poj1979
- poj1979
- poj1979
- poj1979
- poj1979
- 【POJ1979】
- poj1979
- POJ1979
- poj1979--DFS
- poj1979 DFS
- dfs--poj1979
- POJ1979题解
- cookie 和session 的区别详解
- 自己动手做J浏览器——基于JAVA和火狐内核(gecko)
- 为什么选择这种技术而不选择另一种技术?
- 第十三次codeforces竞技结束 #265 Div 2
- myeclipse自动生成hibernate的Mapping和po文件
- poj1979
- 使用mmap和posix semaphores做进程间通信与同步
- CPP -- W3 类和对象进阶
- Block块笔记
- Java笔试题集锦
- 网页结构基础1
- JAVA 基础(一)
- IT忍者神龟之Database Link详解
- 自己的stackoverflow的账号信息