题目:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=7&page=show_problem&problem=513
说明:用DFS求连通块,理解了之后挺简单的。
#include<cstdio> #include<cstring> #include<iostream> using namespace std; char s[110][110]; int vis[110][110]; int M,N,cnt; void dfs(int i,int j){ vis[i][j]=1; for(int a=-1;a<=1;a++){ for(int b=-1;b<=1;b++){ if(a!=0 || b!=0){ if(s[i+a][j+b]=='@' && vis[i+a][j+b]==0) dfs(i+a,j+b); } } } } int main(){ while(cin >> M >> N){ if(M==0&&N==0) break; for(int i=0;i<M;i++) scanf("%s",&s[i]); memset(vis,0,sizeof(vis)); cnt=0; for(int i=0;i<M;i++){ for(int j=0;j<N;j++){ if(s[i][j]=='@' && vis[i][j]==0) { cnt++; dfs(i,j); } } } cout << cnt << endl; } return 0; }
汝佳提供的代码:
// UVa572 Oil Deposits // Rujia Liu // 题意:输入一个字符矩阵,统计字符@组成多少个四连块 #include<cstdio> #include<cstring> const int maxn = 100 + 5; char pic[maxn][maxn]; int m, n, idx[maxn][maxn]; void dfs(int r, int c, int id) { if(r < 0 || r >= m || c < 0 || c >= n) return; if(idx[r][c] > 0 || pic[r][c] != '@') return; idx[r][c] = id; for(int dr = -1; dr <= 1; dr++) for(int dc = -1; dc <= 1; dc++) if(dr != 0 || dc != 0) dfs(r+dr, c+dc, id); } int main() { while(scanf("%d%d", &m, &n) == 2 && m && n) { for(int i = 0; i < m; i++) scanf("%s", pic[i]); memset(idx, 0, sizeof(idx)); int cnt = 0; for(int i = 0; i < m; i++) for(int j = 0; j < n; j++) if(idx[i][j] == 0 && pic[i][j] == '@') dfs(i, j, ++cnt); printf("%d\n", cnt); } return 0; }