题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1198
dfs:
#include<cstdio>//hdu1198 dfs
#include<cstring>
int well[11][5] = { {1,1,1,0,0},{1,0,1,1,0},{0,1,1,0,1},{0,0,1,1,1},{1,0,1,0,1},{0,1,1,1,0},{1,1,1,1,0},{1,1,1,0,1},{0,1,1,1,1},{1,0,1,1,1},{1,1,1,1,1}};
const int ca[5][2] = {0,1,1,0,1,1,1,2,2,1}, mov[4][2] = {1,0,-1,0,0,1,0,-1};
int m,n,sum, map[550][550],id[550][550];
void build(int set, int x, int y)
{
for(int k = 0; k<5; k++)
{
if(well[set][k])
{
map[3*x+ca[k][0]][3*y+ca[k][1]] = 1;
}
}
}
void dfs(int x, int y, int iden)
{
id[x][y] = iden;
for(int i = 0; i<4; i++)
{
int xx = x + mov[i][0], yy = y + mov[i][1];
if(xx<0 || xx>3*m-1 || yy<0 || yy>3*n-1 ) continue;
if(map[xx][yy] && !id[xx][yy])
dfs(xx,yy,iden);
}
}
int main()
{
char init[5005];
while(scanf("%d%d",&m,&n) && (m!=-1 || n!=-1 ))
{
memset(map,0,sizeof(map));
memset(id,0,sizeof(id));
for(int i = 0; i<m; i++)
{
scanf("%s",init);
for(int j = 0; j<n; j++)
build(init[j]-65,i,j);
}
sum = 0;
for(int i = 0; i<=3*m-1; i++)
{
for(int j = 0; j<=3*n-1; j++)
{
//printf("%d ",map[i][j]);
if(map[i][j] && !id[i][j])
{
dfs(i,j,++sum);
}
}
//putchar('\n');
}
printf("%d\n",sum);
}
return 0;
}
并查集:
#include<cstdio>//hdu1198 并查集 每个格子的标号为i*n+j,初始father也为i*n+j,以次来作为合并集合的依据
#include<cstring>
int n,m,father[55][55];
char map[55][55];
char type[11][5]={"1010","1001","0110","0101","1100","0011",
"1011","1110","0111","1101","1111"};
int find(int x)
{
return (x==father[x/n][x%n])?x:find(father[x/n][x%n]);
}
int unit(int a, int b)
{
a = find(a);
b = find(b);
if(a!=b)//合并集合
father[a/n][a%n] = b;
}
int judge(int i, int j)
{ //判断是否与左边相连
if(j>0 && type[map[i][j]-'A'][2]=='1' && type[map[i][j-1]-'A'][3]=='1')
unit(i*n+j,i*n+j-1);
//判断是否与上边相连
if(i>0 && type[map[i][j]-'A'][0]=='1' && type[map[i-1][j]-'A'][1]=='1')
unit(i*n+j,(i-1)*n+j);
}
int main()
{
while(scanf("%d%d",&m,&n) && (m!=-1 || n!=-1))
{
for(int i = 0; i<m; i++)
{
scanf("%s",&map[i]);
for(int j = 0; j<n; j++)
{ //初始化为自己
father[i][j] = i*n+j;
}
}
//并查集过程
for(int i = 0; i<m; i++)
for(int j = 0; j<n; j++)
judge(i,j);
int sum = 0;
for(int i = 0; i<m; i++)
for(int j = 0; j<n; j++)//看看有多少个“根节点”
if(father[i][j]==i*n+j) sum++;
printf("%d\n",sum);
}
}