HDU1198 water pipe and set farm irrigation

Benny has a spacious farm land to irrigate. The farm land was a rectangle, and is divided to a lot of SAMLL squares. Water pipes is placed in these squares. Different Square has a Different type of pipe. There is one types of pipes, which is marked from A to K, as Figure 1 shows.

Farm Irrigation
Figure 1

Benny has a map of his farm, which is an array of marks denoting the distribution of water in the over the pipes farm. For example, if he has a map

Adc
Fjk
IHE

Then the water pipes is distributed like


Figure 2

Several Wellsprings is 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 would have a good harvest in autumn.

Now Benny wants to know on least how many wellsprings should is found to has the whole farm land irrigated. Can you help him?

Note:in The above example, at least 3 wellsprings is needed, as those red points in Figure 2 show.

Inputthere is several test cases! The first line contains is 2 integers m and N, then M lines follow. In each of these lines, there is N characters, in the range of ' A ' to ' K ', denoting the type of water pipe over the Corre Sponding Square. A negative m or N denotes the end of input, else you can assume 1 <= m, N <= 50.

Outputfor each test case, output in one line the least number of wellsprings needed.

Sample Input2 2 DK HF 3 3 ADC fjk IHE-1-1

Sample Output2 3

Authorzheng, Lu

Sourcezhejiang University Local Contest 2005

RECOMMENDIGNATIUS.L problem-solving ideas and check the classification problems, the division of the heap, equals itself is the total number of

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
using namespace Std;
Char a[55][55];
int pd[2505];
struct node
{
int l;
int R;
int u;
int D;
int num;
}QUE[55][55];
int get (int x)
{
if (pd[x]!=x)
Pd[x]=get (Pd[x]);
return pd[x];
}
void find (int x,int y)
{
Pd[get (y)]=get (x);
}
int main ()
{
int m,n;
while (scanf ("%d%d", &m,&n)!=eof)
{
if (m==-1&&n==-1)
Break
if (m==1&&n==1)
{
printf ("1\n");
Continue
}
memset (Pd,0,sizeof (PD));
memset (A,0,sizeof (a));
int n=n*m;
for (int i1=1;i1<=n;i1++)
PD[I1]=I1;
GetChar ();
for (int i2=0;i2<m;i2++)
{
for (int j2=0;j2<n;j2++)
scanf ("%c", &a[i2][j2]);
GetChar ();
}
int h=1;
for (int i=0;i<m;i++)
for (int j=0;j<n;j++)
{
if (a[i][j]== ' a ') {que[i][j].l=1;que[i][j].r=0;que[i][j].u=1;que[i][j].d=0;}
else if (a[i][j]== ' B ') {que[i][j].l=0;que[i][j].r=1;que[i][j].u=1;que[i][j].d=0;}
else if (a[i][j]== ' C ') {que[i][j].l=1;que[i][j].r=0;que[i][j].u=0;que[i][j].d=1;}
else if (a[i][j]== ' D ') {que[i][j].l=0;que[i][j].r=1;que[i][j].u=0;que[i][j].d=1;}
else if (a[i][j]== ' E ') {que[i][j].l=0;que[i][j].r=0;que[i][j].u=1;que[i][j].d=1;}
else if (a[i][j]== ' F ') {que[i][j].l=1;que[i][j].r=1;que[i][j].u=0;que[i][j].d=0;}
else if (a[i][j]== ' G ') {que[i][j].l=1;que[i][j].r=1;que[i][j].u=1;que[i][j].d=0;}
else if (a[i][j]== ' H ') {que[i][j].l=1;que[i][j].r=0;que[i][j].u=1;que[i][j].d=1;}
else if (a[i][j]== ' I ') {que[i][j].l=1;que[i][j].r=1;que[i][j].u=0;que[i][j].d=1;}
else if (a[i][j]== ' J ') {que[i][j].l=0;que[i][j].r=1;que[i][j].u=1;que[i][j].d=1;}
else {que[i][j].l=1;que[i][j].r=1;que[i][j].u=1;que[i][j].d=1;}
Que[i][j].num=h;
h++;
}
for (int i3=0;i3<m;i3++)
for (int j3=0;j3<n;j3++)
{
if (j3<n-1)
{
if (que[i3][j3].r==1&&que[i3][j3+1].l==1)
Find (Que[i3][j3].num,que[i3][j3+1].num);
}

if (i3<m-1)
{
if (que[i3][j3].d==1&&que[i3+1][j3].u==1)
Find (Que[i3][j3].num,que[i3+1][j3].num);
}
}
int sum=0;
for (int i4=1;i4<=n;i4++)
{
if (PD[I4]==I4)
sum++;
}
printf ("%d\n", sum);
}
return 0;
}

HDU1198 water pipe and set farm irrigation