HDU1241 DFS/BFS 求连通块
0)
遍历所有点，求出连通块的数量，每到一块油田，就采用DFS或者BFS消除该油田以及附近连通的油田。
注意（易错）:
①二维数组mat[i][j]，i是第几行，j是第几列，所以搜到一点再搜该点周围八个方位(+dir[i][j])时，以及确定是否超出地图范围时，注意i是纵向加减，j是横向加减，且(0,0)原点在左上角，向右向下坐标的值都是不断递增。
②scanf("%c",&mat[i][j]); ， scanf读入%c的字符时，会出错！可改成cin>>，或者scanf("%s",&mat[i])，如果数组第二维从1开始就scanf("%s",&mat[i]+1)，scanf依然比cin快。（见如下BFS代码中的使用）
1)
DFS
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
const int maxn=105;
char mat[maxn][maxn];
int dir[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{-1,1},{1,-1},{-1,-1}};
int m,n;
int sum;
void Dfs(int x,int y){
int tx;
int ty;
for(int i=0;i<8;i++){
tx=x+dir[i][0];
ty=y+dir[i][1];
if(1<=tx&&tx<=m&&1<=ty&&ty<=n){//注意！对于二维数组来说，x行y列，意味着x是纵向坐标，y是横向坐标
if(mat[tx][ty]=='@'){
mat[tx][ty]='*';
Dfs(tx,ty);
}
}
}
}
int main()
{
while(scanf("%d%d",&m,&n)){
if(m==0&&n==0){
break;
}
//memset(mat,0,sizeof(mat));
sum=0;
for(int i=1;i<=m;i++){
for(int j=1;j<=n;j++){
//scanf("%s",mat[i]+1);
cin>>mat[i][j];
//scanf("%c",&mat[i][j]);
}
}
for(int i=1;i<=m;i++){
for(int j=1;j<=n;j++){
if(mat[i][j]=='@'){
sum++;
mat[i][j]='*';
Dfs(i,j); //深搜，消除同一连通块的所有油田标记
}
}
}
cout<<sum<<endl;
}
return 0;
}
BFS
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
const int maxn=105;
char mat[maxn][maxn];
int dir[8][2]={{1,0},{-1,0},{0,1},{0,-1},{1,1},{-1,1},{1,-1},{-1,-1}};
int m,n;
int sum;
struct Node{
int x;
int y;
};
void Bfs(int x,int y){
queue <Node> q;
Node node;
node.x=x;
node.y=y;
q.push(node);
while(!q.empty()){
Node cur=q.front();
Node next;
q.pop();
for(int i=0;i<8;i++){
next.x=cur.x+dir[i][0];
next.y=cur.y+dir[i][1];
if(mat[next.x][next.y]=='@'){
mat[next.x][next.y]='*';
q.push(next);
}
}
}
}
int main()
{
while(scanf("%d %d",&m,&n)){
if(m==0&&n==0){
break;
}
memset(mat,0,sizeof(mat));
sum=0;
int cur=1;
for(int i=1;i<=m;i++){
//for(int j=1;j<=n;j++){
scanf("%s",mat[i]+1);//因为第二维也都是从1开始到n，所以+1开始
//cin>>mat[i][j];
//scanf("%c",&mat[i][j]);
//}
}
for(int i=1;i<=m;i++){
for(int j=1;j<=n;j++){
if(mat[i][j]=='@'){
sum++;
mat[i][j]='*';
Bfs(i,j); //广搜，消除同一连通块的所有油田标记
}
}
}
cout<<sum<<endl;
}
return 0;
}
2）
Description
The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.
Input
The input file contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*', representing the absence of oil, or `@', representing an oil pocket.
Output
For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.
Sample Input
1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5
****@
*@@*@
*@**@
@@@*@
@@**@
0 0
Sample Output
0
1
2
2