Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2492 Accepted Submission(s): 1107
Benny has a map of his farm, which is an array of marks denoting the distribution of water pipes over the whole farm. For example, if he has a map
ADC
FJK
IHE
then the water pipes are distributed like
Several wellsprings are found in the center of some squares, so water can flow along the pipes from one square to another. If water flow crosses one square, the whole farm land in this square is irrigated and will have a good harvest in autumn.
Now Benny wants to know at least how many wellsprings should be found to have the whole farm land irrigated. Can you help him?
Note: In the above example, at least 3 wellsprings are needed, as those red points in Figure 2 show.
//做的最郁闷的一次并查集了，看到这个set()函数就知道哪郁闷了
//
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
bool h1[100][100];
bool h2[100][100];
void set()
{
h1['B']['A']=h1['B']['C']=h1['B']['F']=h1['B']['G']=h1['B']['H']=h1['B']['I']=h1['B']['K']=1;
h1['D']['A']=h1['D']['C']=h1['D']['F']=h1['D']['G']=h1['D']['H']=h1['D']['I']=h1['D']['K']=1;
h1['F']['A']=h1['F']['C']=h1['F']['F']=h1['F']['G']=h1['F']['H']=h1['F']['I']=h1['F']['K']=1;
h1['G']['A']=h1['G']['C']=h1['G']['F']=h1['G']['G']=h1['G']['H']=h1['G']['I']=h1['G']['K']=1;
h1['I']['A']=h1['I']['C']=h1['I']['F']=h1['I']['G']=h1['I']['H']=h1['I']['I']=h1['I']['K']=1;
h1['J']['A']=h1['J']['C']=h1['J']['F']=h1['J']['G']=h1['J']['H']=h1['J']['I']=h1['J']['K']=1;
h1['K']['A']=h1['K']['C']=h1['K']['F']=h1['K']['G']=h1['K']['H']=h1['K']['I']=h1['K']['K']=1;
h2['C']['A']=h2['C']['B']=h2['C']['E']=h2['C']['G']=h2['C']['H']=h2['C']['J']=h2['C']['K']=1;
h2['D']['A']=h2['D']['B']=h2['D']['E']=h2['D']['G']=h2['D']['H']=h2['D']['J']=h2['D']['K']=1;
h2['E']['A']=h2['E']['B']=h2['E']['E']=h2['E']['G']=h2['E']['H']=h2['E']['J']=h2['E']['K']=1;
h2['H']['A']=h2['H']['B']=h2['H']['E']=h2['H']['G']=h2['H']['H']=h2['H']['J']=h2['H']['K']=1;
h2['I']['A']=h2['I']['B']=h2['I']['E']=h2['I']['G']=h2['I']['H']=h2['I']['J']=h2['I']['K']=1;
h2['J']['A']=h2['J']['B']=h2['J']['E']=h2['J']['G']=h2['J']['H']=h2['J']['J']=h2['J']['K']=1;
h2['K']['A']=h2['K']['B']=h2['K']['E']=h2['K']['G']=h2['K']['H']=h2['K']['J']=h2['K']['K']=1;
}
int n,m,t;
int f[2503],r[2503];
int find_f(int x)
{
if(x!=f[x])
{
return f[x]=find_f(f[x]);
}
return x;
}
void union_set(int x,int y)
{
x=find_f(x);
y=find_f(y);
if(x==y) return;
t--;
if(r[x]>r[y])
{
f[y]=x;
}
else if(r[x]<r[y])
{
f[x]=y;
}
else
{
f[y]=x;
r[x]++;
}
}
char s[66][66];
int main()
{
int i,j,a,b;
set();
while(scanf("%d%d",&n,&m),n>0&&m>0)
{
memset(s,0,sizeof(s));//开始忘记这个、WA了
for(i=1;i<=n;i++)
scanf("%s",s[i]+1);
t=n*m;
for(i=1;i<=t;i++)
f[i]=i,r[i]=0;
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
{
if(h1[s[i][j]][s[i][j+1]])
{
a=(i-1)*m+j;
b=a+1;
union_set(a,b);
}
if(h2[s[i][j]][s[i+1][j]])
{
a=(i-1)*m+j;
b=i*m+j; //这个也写错了次，Wa原因之一呀
union_set(a,b);
}
}
printf("%d\n",t);
}
return 0;
}
