HDU 1312 Red and Black（BFS，DFS）

Red and Black

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

Total Submission(s): 16081 Accepted Submission(s): 9912



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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)

[align=left]Output[/align]
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).

[align=left]Sample Input[/align]

6 9
....#.
.....#
......
......
......
......
......
#@...#
.#..#.
11 9
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
0 0

[align=left]Sample Output[/align]

45
59
6
13

[align=left]Source[/align]
Asia 2004, Ehime (Japan), Japan Domestic

题解：经典的搜索（bfs，dfs也可以）

TLE代码.....:搜得太深了。。。

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<algorithm>
typedef long long LL;
using namespace std;
int w,h;
char z[21][21];
int dfs(int i,int j){
if(i<1||j>h||j<1||j>w)return 0;

if(z[i][j]!='#')
{
z[i][j]='#';
return 1+dfs(i-1,j)+dfs(i,j-1)+dfs(i+1,j)+dfs(i,j+1);  //TLE....
}
else return 0;
}
int main()
{

while(cin>>w>>h){
if(w==0&&h==0)break;

for(int i=1;i<=h;i++)
for(int j=1;j<=w;j++)
cin>>z[i][j];
for(int i=1;i<=h;i++){
for(int j=1;j<=w;j++)
if(z[i][j]=='@')
cout<<dfs(i,j)<<endl;
}
}
return 0;
}

AC代码：（BFS）

#include <stdio.h>
#include <queue>
#include <string.h>
#include <algorithm>
using namespace std;
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}}; //扩展的四个方向
char map[33][33];
int vis[33][33];
int n,m;
int sum;
struct node
{
int x,y;
}
f[333];
#define inf 0xfffffff
void bfs(int x,int y)
{
int i;
queue<node>q;
node st,ed;
st.x=x;
st.y=y;
q.push(st);
while(!q.empty())  //队列非空
{
st=q.front();
q.pop();
for(i=0;i<4;i++)
{
ed.x=st.x+dir[i][0];
ed.y=st.y+dir[i][1];
if(ed.x>=n ||ed.y>=m ||ed.x<0 ||ed.y<0 ||map[ed.x][ed.y]=='#' ||map[ed.x][ed.y]=='@') //判断越界
continue;
map[ed.x][ed.y]='@';  //标志已经遍历过的
sum++;
q.push(ed);
}
}
}
int main()
{
int i,j,x,y;
while(scanf("%d%d",&m,&n)!=EOF&&n!=0 &&m!=0)
{
for(i=0;i<n;i++)
scanf("%s",map[i]);
for(i=0;i<n;i++)
for(j=0;j<m;j++)
{
if(map[i][j]=='@')  //寻找起始点
{
x=i;
y=j;
}
}
sum=1;
bfs(x,y);
printf("%d\n",sum);
}
return 0;
}
//广搜

AC代码：（DFS）

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int dir[4][2]={{1,0},{0,1},{-1,0},{0,-1}};
char map[33][33];
int vis[33][33];
int n,m;
int sum;
#define inf 0xfffffff
void dfs(int x,int y)
{
int sx,sy,i;
sum++;
for(i=0;i<4;i++)
{
sx=x+dir[i][0];
sy=y+dir[i][1];
if(sx>=n ||sy>=m ||sx<0 ||sy<0 ||map[sx][sy]=='#' ||map[sx][sy]=='@')
continue;
map[sx][sy]='@';
dfs(sx,sy);
}
}
int main()
{
int i,j,x,y;
while(scanf("%d%d",&m,&n)!=EOF&&n!=0 &&m!=0)
{
for(i=0;i<n;i++)
scanf("%s",map[i]);
for(i=0;i<n;i++)
for(j=0;j<m;j++)
{
if(map[i][j]=='@')
{
x=i;
y=j;
}
}
sum=0;
dfs(x,y);
printf("%d\n",sum);
}
return 0;
}
//深搜