Time Limit:
MS (Java/Others)
Memory Limit:
K (Java/Others)Total Submission(s): 30156
Accepted Submission(s): 17445
Problem Description
Sample Input
Sample Output
DFS求连通块，DFS经典应用题。。。
#include &bits/stdc++.h&
#define MAX_N 110
char maze[MAX_N][MAX_N];
void solve();
void dfs(int x, int y);
int main()
while (cin && m && n && m && n) {
for (int i = 0; i & i++) {
while(getchar() != '\n')
for (int j = 0; j & j++) {
scanf("%c", &maze[i][j]);
void solve() {
int res = 0;
for (int i = 0; i & R; i++) {
for (int j = 0; j & C; j++) {
if (maze[i][j] == '@') {
dfs(i, j);
cout && res &&
//int dx[] = {-1, 0, 0, 1};
//int dy[] = { 0, 1,-1, 0};
void dfs(int x, int y) {
maze[x][y] = '*';
for (int dx = -1; dx &= 1; dx++) {
for (int dy = -1; dy &= 1; dy++) {
int nx = dx + x, ny = dy +
if (nx &= 0 && nx & R && ny &= 0 && ny & C && maze[nx][ny] == '@') {
dfs(nx, ny);
hdu 1241 最基本的DFS题目
Oil Deposits (搜索)
poj1562 Oil Deposits(简单的深搜)
PKU 1562/HDU 1241
Oil Deposits(原油有多少块区域---BFS,DFS)
没有更多推荐了，这题目相对来说还算是比较简单，深搜......
include&iostream&
#include&stdio.h&
#include&cstring&
char abss[100][100];
int dx[9]={1, 1,0,-1,-1, -1, 0,1};//dx,dy代表方向
int dy[9]={0,-1,-1,-1,0,1, 1, 1};
void fun(int x,int y)
abss[x][y]='*';
for(int i=0;i&8;i++)
int anx=x+dx[i];
int any=y+dy[i];
if(anx&=1&&anx&=a&&any&=1&&any&=b&&abss[anx][any]=='@')
fun(anx,any);
int main()
while(cin&&a&&b&&a&&b)
int count=0;
for(int i=1;i&=a;i++)
for(int j=1;j&=b;j++)cin&&abss[i][j];
for(int i=1;i&=a;i++)
for(int j=1;j&=b;j++)
if(abss[i][j]=='@')
cout&&count&&
没有更多推荐了，Sina Visitor System}