Farm Irrigation HDU

题目：

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
Sample Output
2
3

题目大意：

有上述11种水管摆放方式，现给你一块n*m的农田，问你需要多少个出水点能灌溉到所有地方。

思路：

并查集求联通快.

代码：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<vector>#include<stack>#include<bitset>#include<cstdlib>#include<cmath>#include<cctype>#include<sstream>#include<set>#include<list>#include<deque>#include<map>#include<queue>using namespace std;typedef long long ll;const double PI = acos(-1.0);const double eps = 1e-6;const int INF = 0x3f3f3f3f;const int maxn = 123456;int T,n,m;int f[3100];char maps[55][55];//A-K向四个方向是否连通 左上右下int dir[11][5]={{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}};int Get_f(int x){    if(f[x]==-1)        return x;    return Get_f(f[x]);}void Merge(int x,int y){    int t1=Get_f(x);    int t2=Get_f(y);    if(t1!=t2)        f[t1]=t2;}int main(){    while(scanf("%d%d",&n,&m)!=EOF!=EOF)    {        if(n==-1&&m==-1)            break;        for(int i=0;i<n;i++)            scanf("%s",&maps[i]);        memset(f,-1,sizeof(f));        for(int i=0;i<n;i++)        {            for(int j=0;j<m;j++)            {                if(i>0&&dir[maps[i][j]-'A'][1]==1&&dir[maps[i-1][j]-'A'][3]==1)//上方可以连接                    Merge(i*m+j,(i-1)*m+j);                if(j>0&&dir[maps[i][j]-'A'][0]==1&&dir[maps[i][j-1]-'A'][2]==1)//左方                    Merge(i*m+j,i*m+j-1);                if(i<n-1&&dir[maps[i][j]-'A'][3]==1&&dir[maps[i+1][j]-'A'][1]==1)//下方                    Merge(i*m+j,(i+1)*m+j);                if(j<m-1&&dir[maps[i][j]-'A'][2]==1&&dir[maps[i][j+1]-'A'][0]==1)//右方                    Merge(i*m+j,i*m+j+1);            }        }        int cnt=0;        for(int i=0;i<n*m;i++)            if(f[i]==-1)                cnt++;        printf("%d\n",cnt);    }    return 0;}