HDU 2952 Counting Sheep(DFS)
Creative as I am, that wasn't going to stop me. I sat down and wrote a computer program that made a grid of characters, where # represents a sheep, while . is grass (or whatever you like, just not sheep). To make the counting a little more interesting, I also decided I wanted to count flocks of sheep instead of single sheep. Two sheep are in the same flock if they share a common side (up, down, right or left). Also, if sheep A is in the same flock as sheep B, and sheep B is in the same flock as sheep C, then sheeps A and C are in the same flock.
Now, I've got a new problem. Though counting these sheep actually helps me fall asleep, I find that it is extremely boring. To solve this, I've decided I need another computer program that does the counting for me. Then I'll be able to just start both these programs before I go to bed, and I'll sleep tight until the morning without any disturbances. I need you to write this program for me.
Each test case begins with a line containing two numbers, H and W, the height and width of the sheep grid. Then follows H lines, each containing W characters (either # or .), describing that part of the grid.
Notes and Constraints
0 < T <= 100
0 < H,W <= 100
4 4
题解:题目描述什么的一大段废话,直接看输入输出,发现是一个经典的DFS连通块问题。
#include <cstdio>
#include <iostream>
#include <string>
#include <sstream>
#include <cstring>
#include <stack>
#include <queue>
#include <algorithm>
#include <cmath>
#include <map>
#define ms(a) memset(a,0,sizeof(a))
#define msp memset(mp,0,sizeof(mp))
#define msv memset(vis,0,sizeof(vis))
using namespace std;
//#define LOCAL
char mp[][];
int n,m;
int ans;
int dir[][]={{,},{,-},{,},{-,}};
void dfs(int x,int y)
{
for(int i=;i<;i++)
{
int nx=x+dir[i][],ny=y+dir[i][];
if(nx>=&&nx<n&&ny>=&&ny<m)
{
if(mp[nx][ny]=='#')
{
mp[nx][ny]='.';
dfs(nx,ny);
}
}
}
return;
}
int main()
{
#ifdef LOCAL
freopen("in.txt", "r", stdin);
#endif // LOCAL
//Start
int N;
cin>>N;
while(N--)
{
cin>>n>>m;
msp,ans=;
for(int i=; i<n; i++)cin>>mp[i];
for(int i=; i<n; i++)
for(int j=; j<m; j++)
if(mp[i][j]=='#')
{
mp[i][j]='.';
ans++;
dfs(i,j);
}
printf("%d\n",ans);
}
return ;
}
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