HDU 1198 Farm Irrigation

Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 250 Accepted Submission(s): 121 

Problem Description

Benny has a spacious farm land to irrigate. The farm land is a rectangle, and is divided into a lot of samll squares. Water pipes are placed in these squares. Different square has a different type of pipe. There are 11 types of pipes, which is marked from A to K, as Figure 1 shows.


Figure 1

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 


Figure 2

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.

 

Input

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

 

Output

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

 

Sample Input

2 2DKHF3 3ADCFJKIHE-1 -1

 

Sample Output

23

 

题目大意：

给出农田，给出农田是由哪几块拼起来的，然后问你有几个连通的。

思路：

标记起来每块农田上下左右的出口，然后依次开始检索，每次只检索右边和下边就行了，这样就能检索完整个图。在判断有几个根。

完美。

AC代码：

#include<iostream>#include<stdio.h>#include<string.h>using namespace std;const int MAX = 5000;int parent[MAX+10];//total[GetParent(a)]是a所在的 group的人数int GetParent(int a){    //获取a的根,并把a的父节点改为根    if( parent[a]!= a)        parent[a] = GetParent(parent[a]);    return parent[a];}void Merge(int a,int b){    int p1 = GetParent(a);    int p2 = GetParent(b);    if( p1 == p2 )        return;    parent[p2] = p1;}int mapp[15][4]= {1,1,0,0,0,1,1,0,1,0,0,1,0,0,1,1,0,1,0,1,1,0,1,0,1,1,1,0,1,1,0,1,1,0,1,1,0,1,1,1,1,1,1,1};bool leftt(int x,int y){    if(mapp[x][2]==1&&mapp[y][0]==1)        return 1;    return 0;}bool upp(int x,int y){    if(mapp[x][3]==1&&mapp[y][1]==1)        return 1;    return 0;}int main(){    int n,m;    char ch[55];    int haha[55][55];    while(scanf("%d%d",&n,&m))    {        if(n<0&&m<0)            break;        for(int i = 0; i < 2550; ++i)        {            parent[i] = i;        }        for(int i=0; i<n; i++)        {            scanf("%s",ch);            for(int j=0; j<m; j++)            {                haha[i][j]=ch[j]-'A';            }        }        int noww,nexx;        for(int i=0; i<n; i++)        {            for(int j=0; j<m; j++)            {                if(j!=m-1&&leftt(haha[i][j],haha[i][j+1]))                {                    Merge(i*m+j,i*(m)+j+1);                }                if(i!=n-1&&upp(haha[i][j],haha[i+1][j]))                {                    Merge(i*(m)+j,(i+1)*(m)+j);                }            }        }        int t=0;        int g[MAX+10];        memset(g,0,sizeof(g));        for(int i=0; i<(n)*(m); i++)              if(GetParent(i)==i)                g[GetParent(i)]++;        for(int i=0; i<(n+1)*(m+1); i++)              if(g[i]>0)                t++;        cout<<t<<endl;    }}