HDU 1312 Red and Black 红与黑 搜索 dfs bfs

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 17114 Accepted Submission(s): 10411

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

45

59

6

13

dfs入门 复制过来的题目描述变得很奇怪。。

#include<iostream>
#include<stdio.h>
#include<cstring>

using namespace std;

const int cX[4]={-1,0,0,1};
const int cY[4]={0,-1,1,0};
const int MAX=20+5;
int W,H,idx[MAX][MAX],cnt;
char pic[MAX][MAX];

void dfs(int x,int y,int id)
{
if (x<0||x>=H||y<0||y>=W)
return;
if (idx[x][y]>0||pic[x][y]=='#')
return;
idx[x][y]=id;
cnt++;
for (int i=0;i<4;i++)
dfs(x+cX[i],y+cY[i],id);
}

int main()
{
while (scanf("%D%d",&W,&H)==2&&W&&H)
{
for (int i=0;i<H;i++)
scanf("%s",pic[i]);
memset(idx,0,sizeof(idx));
cnt=0;
for (int i=0;i<H;i++)
for (int j=0;j<W;j++)
if (pic[i][j]=='@')
dfs(i,j,1);
printf("%d\n",cnt);
}
return 0;
}

bfs解法

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<queue>
#include<algorithm>
#include<functional>
#include<vector>
#include<iomanip>

using namespace std;

const int dX[4]={-1,0,0,1};
const int dY[4]={0,-1,1,0};
const int MAX=20+5;
int W,H,idx[MAX][MAX],cnt;
char pic[MAX][MAX];

bool check(int x,int y)
{
if (x<0||x>=H||y<0||y>=W)
return false;
if (idx[x][y]>0||pic[x][y]=='#')
return false;
return true;
}

void bfs(int x,int y)
{
pair<int,int> p=make_pair(x,y);
queue<pair<int,int>> Q;
idx[x][y]=1;
cnt++;
Q.push(p);
while (!Q.empty())
{
pair<int,int> u=Q.front();
Q.pop();
int cx=u.first;
int cy=u.second;
for (int i=0;i<4;i++)
{
int nx=cx+dX[i];
int ny=cy+dY[i];
if (check(nx,ny))
{
cnt++;
idx[nx][ny]=1;
Q.push(make_pair(nx,ny));
}
}
}
}

int main()
{
while (scanf("%D%d",&W,&H)==2&&W&&H)
{
for (int i=0;i<H;i++)
scanf("%s",pic[i]);
memset(idx,0,sizeof(idx));
cnt=0;
for (int i=0;i<H;i++)
for (int j=0;j<W;j++)
if (pic[i][j]=='@')
bfs(i,j);
printf("%d\n",cnt);
}
return 0;
}