Farm Irrigation（BFS）

Farm Irrigation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 263 Accepted Submission(s): 133

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 2
DK
HF

3 3
ADC
FJK
IHE

-1 -1
/*
解题思路
本题一开始想用并查集 后来没有思路 然后就直接采用了广搜 每进行广搜就需要num++； 还有一点值得注意是 移动的方向要与dir方向一致这样便于计算
*/

#include<iostream>#include<algorithm>#include<stdio.h>#include<string>#include<string.h>#include<queue>using namespace std;typedef struct LNode{    int x;    int y;}node;//上下左右int dir[11][4]={1,0,1,0,1,0,0,1,0,1,1,0,0,1,0,1,1,1,0,0,0,0,1,1,1,0,1,1,1,1,1,0,0,1,1,1,1,1,0,1,1,1,1,1};int dd[4][2]={-1,0,1,0,0,-1,0,1};//上下左右int h,w;char mp[52][52];int mark[52][52];void bfs(int x,int y){    node now;    queue<LNode> q;    now.x = x;    now.y = y;    mark[x][y] = 1;    q.push(now);    while(!q.empty())    {        now = q.front();        q.pop();        for(int i =0;i<4;i++)        {            int xx = now.x + dd[i][0];            int yy = now.y + dd[i][1];            if(xx>=0&&yy>=0&&xx<h&&yy<w&&!mark[xx][yy])            {                if(i==0||i==1)//判断上下是否联通                {                    if(dir[mp[now.x][now.y]-'A'][i]&&dir[mp[xx][yy]-'A'][1-i])                    {                        mark[xx][yy] =1;                        node nxt;                        nxt.x = xx;                        nxt.y = yy;                        q.push(nxt);                    }                }                else                {                    if(dir[mp[now.x][now.y]-'A'][i]&&dir[mp[xx][yy]-'A'][5-i])                    {                        mark[xx][yy] =1;                        node nxt;                        nxt.x = xx;                        nxt.y = yy;                        q.push(nxt);                    }                }            }        }    }}int main(){    while(cin>>h>>w&&h>=1&&w>=1)    {        for(int i = 0;i<h;i++)        {            cin>>mp[i];        }        int num =0;        memset(mark,0,sizeof(mark));        for(int i =0;i<h;i++)        {            for(int j =0;j<w;j++)            {                if(mark[i][j])continue;                bfs(i,j);                num++;            }        }        cout<<num<<endl;    }    return 0;}