HDU 1312Red and Black(简单搜索 bfs或dfs)

Red and BlackTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6908 Accepted Submission(s): 4387


Problem DescriptionThere 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.

InputThe 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)

OutputFor 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

SourceAsia 2004, Ehime (Japan), Japan Domestic


题目意思就不多说了，很容易理解，@为起点，可以上下左右四个方向扩展，只能走点，问可以到达多少个位置。

DFS AC代码：

#include<iostream>#include<cstring>#include<string>#include<cstdio>#include<queue>using namespace std;int r,c;int dir[4][2]={{0,-1},{0,1},{1,0},{-1,0}};char s[25][25];int visi[25][25];void bfs(int t){    queue<int> mq;    mq.push(t);    while(!mq.empty())    {        int p=mq.front();        mq.pop();        int ta=p/c,tb=p%c;        for(int i=0;i<4;i++)        {            int tta=dir[i][0]+ta,ttb=dir[i][1]+tb;            if(tta>=0&&tta<r&&ttb>=0&&ttb<c)            {                if(!visi[tta][ttb]&&s[tta][ttb]=='.')                {                    visi[tta][ttb]=1;                    mq.push(tta*c+ttb);                }            }        }    }}int main(){    int i,j;    while(scanf("%d%d",&c,&r))    {        if(!r&&!c) break;        memset(visi,0,sizeof(visi));        for(i=0;i<r;i++)            scanf("%s",s[i]);        int a,b;  //把起点的位置找到        for(i=0;i<r;i++)            for(j=0;j<c;j++)                if(s[i][j]=='@')                {                    a=i,b=j;                    break;                }        visi[a][b]=1;        bfs(a*c+b);        int res=0;  //统计访问的个数        for(i=0;i<r;i++)            for(j=0;j<c;j++)                if(visi[i][j]==1)                    res++;        printf("%d\n",res);    }    return 0;}//15MS