Total Submission(s): 1504 Accepted Submission(s): 981
![](http://acm.hdu.edu.cn/data/images/con211-1007-1.jpg)
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
2 4 4 #.#. .#.# #.## .#.# 3 5 ###.# ..#.. #.###
6 3
解题思路:本题是简单的广度优先搜索题,可视为图的连通性查找,找到大图有几个连通的小图即可。
数羊——>数羊群,遍历整个草原,一旦的发现羊,就把整群都标记了,避免重数。然后,继续遍历,直到整个草原遍历完。
#include<cstdio> #include<cstring> #include<queue> using namespace std; int n,m; char map[100][100]; int sum; int dir[4][2]={1,0,0,1,0,-1,-1,0}; struct node { int x; int y; }; int inmap(int x,int y) { if(x>=0&&x<n&&y>=0&&y<m) return true; return false; } void bfs(int x,int y) { int i; node u,v; queue<node> q; u.x=x; u.y=y; q.push(u); while(!q.empty()) { u=q.front(); q.pop(); for(i=0;i<4;i++) { v.x=u.x+dir[i][0]; v.y=u.y+dir[i][1]; if(inmap(v.x,v.y)&&map[v.x][v.y]=='#') //这里有一只没有数过的羊 { map[v.x][v.y]='*'; //这只羊已经数过了,下次不数它 q.push(v); } } } } int main() { int i,j; int t; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); sum=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]=='#') //发现羊 { sum++; bfs(i,j); //找到整群羊,并标记,避免重数 } } } printf("%d\n",sum); } return 0; }