HDU 1312 Red and Black 【DPS】

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13487    Accepted Submission(s): 8359

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

 

题意：

从@开始走 #是墙 计算@能走的位置 ‘.’表示可以走1地方

注意：

1：本题没有用到回溯 因为可以返回到走过的位置 但是不再增加步数

2：31行getchar（）很重要

3：map 是地图 step是走下一步的方向

 

#include<stdio.h>#include<string.h>int a,b,ax,ay,nx,ny,s,step[4][2]= {{1,0},{-1,0},{0,1},{0,-1}};char map[25][25];void DFS(int x,int y){    //if(x<=0||x>b||y<=0||y>a) return ;//并没什么用    //if(map[x][y]=='#') return ;//之前这里的问题找了好久 一直没发现为什么 在被标记#后再DPS 判断#后直接return，无法进行下一步搜索    for(int i=0; i<=3; i++)    {        // printf("%d %d  x\n",x,y);        nx=x+step[i][0];        ny=y+step[i][1];        //if(nx<=0||nx>b||ny<=0||ny>a||map[nx][ny]=='#'||map[nx][ny]=='O') continue; 不符合的条件        if(map[nx][ny]=='.')//符合要求        {            s++;            map[nx][ny]='#';            DFS(nx,ny);        }    }}int main (void){    int i,ii;    while(~scanf("%d%d",&a,&b),a||b)    {        s=0;        memset(map,'0',sizeof(map));        getchar();        for(i=1; i<=b; i++)        {            for(ii=1; ii<=a; ii++)            {                scanf("%c",&map[i][ii]);                if(map[i][ii]=='@')                {                    ax=i;                    ay=ii;                }            }            getchar();        }        s=0;        DFS(ax,ay);        printf("%d\n",s+1);//没有计算起点 所以+1    }    return 0;}