HDU 1312 Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12130    Accepted Submission(s): 7550

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

4559613

 

Source

Asia 2004, Ehime (Japan), Japan Domestic

 

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这是我真正接触ACM的第一题，也是大一暑假的第一题，但是快一年了，我的水平几乎没有什么本质的提升，我真的浪费了一年，起码在ACM方面，我很后悔！

今天算是复习，把DFS和BFS两种代码都贴一下
首先是DFS

#include<stdio.h>#include<queue>#include<iostream>using namespace std;char mat[111][111];int m,n,cnt;int dx[]={1,-1,0,0};int dy[]={0,0,1,-1};void dfs(int x,int y){    if(mat[x][y]=='#'||x<1||y<1||x>n||y>m)return;    mat[x][y]='#';    cnt++;    for(int i=0;i<4;i++)    {        dfs(x+dx[i],y+dy[i]);    }}int main(){    int sx,sy;    while(~scanf("%d%d",&m,&n),(m+n))    {        cnt=0;        for(int i=1;i<=n;i++)            for(int j=1;j<=m;j++)        {            scanf(" %c",&mat[i][j]);            if(mat[i][j]=='@')sx=i,sy=j;        }        dfs(sx,sy);        cout<<cnt<<endl;    }}

然后是BFS

#include<stdio.h>#include<queue>#include<iostream>using namespace std;struct point{    int x,y;}st;queue<point>q;char mat[111][111];int m,n,cnt;int dx[]={1,-1,0,0};int dy[]={0,0,1,-1};void bfs(int x,int y){    while(!q.empty())q.pop();    q.push(st);    mat[st.x][st.y]='#';    while(!q.empty())    {        st=q.front();        q.pop();        cnt++;        for(int i=0;i<4;i++)        {            point next=st;            next.x+=dx[i],next.y+=dy[i];            if(mat[next.x][next.y]=='#'||next.x<1||next.y<1||next.x>n||next.y>m)continue;            q.push(next);            mat[next.x][next.y]='#';        }    }}int main(){    int sx,sy;    while(~scanf("%d%d",&m,&n),(m+n))    {        cnt=0;        for(int i=1;i<=n;i++)            for(int j=1;j<=m;j++)        {            scanf(" %c",&mat[i][j]);            if(mat[i][j]=='@')st.x=i,st.y=j;        }        bfs(sx,sy);        cout<<cnt<<endl;    }return 0;}