hdu 1198 农田灌溉 并查集4

Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6539    Accepted Submission(s): 2827

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

#include <iostream>using namespace std;int fa[3000];       //注意不能开小,大于50*50int findfa(int x){    /*if(x==fa[x])      return x;    else return fa[x]=findfa(fa[x]); */    while(fa[x]!=x)        x=fa[x];    return x;}int unionf(int x,int y){    int fx=findfa(x);    int fy=findfa(y);    fa[fx]=fy;    return 0;}int main(){    int a[12][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}};    //将每种图形储存在数组中    int m,n,i,j,k,c;    while(cin>>m>>n)    {        if(m==-1&&n==-1)      //不用连等            break;        int b[60][60];        char b1;        for(i=0; i<m; i++)            for(j=0; j<n; j++)            {                cin>>b1;                b[i][j]=b1-'A';            }        c=m*n;        for(i=1; i<=c; i++)            fa[i]=i;        for(i=0; i<m-1; i++)          //比较前m-1行，n-1列        {            for(j=0; j<n-1; j++)            {                if(a[b[i][j]][1]&&a[b[i+1][j]][0])                    unionf(i*n+j+1,(i+1)*n+j+1);//a[i][j]的下边，a[i+1][j]的上边都为1，可合并                if(a[b[i][j]][3]&&a[b[i][j+1]][2])//a[i][j]的右边，a[i][j+1]的左边都为1，可合并                    unionf(i*n+j+1,i*n+j+2);            }        }        for(i=0; i<m-1; i++) //j=n-1时，即比较最后一列上下块是否可以合并，因为比较上下时有i+1，            //右下角块会超出范围        {            if(a[b[i][n-1]][1]&&a[b[i+1][n-1]][0])                unionf(i*n+n,(i+1)*n+n);        }        for(j=0; j<n-1; j++) //i=m-1时，即比较最后一行左右块是否可以合并        {            if(a[b[m-1][j]][3]&&a[b[m-1][j+1]][2])                unionf((m-1)*n+j+1,(m-1)*n+j+2);        }        k=0;        for(i=1; i<=c; i++)        {            if(fa[i]==i)                k++;        }        cout<<k<<endl;    }    return 0;}

优化：

#include <iostream>using namespace std;int fa[3000];       //注意不能开小,大于51*51int findfa(int x){    /*if(x==fa[x])      return x;    else return fa[x]=findfa(fa[x]); */    while(fa[x]!=x)        x=fa[x];    return x;}int unionf(int x,int y){    int fx=findfa(x);    int fy=findfa(y);    fa[fx]=fy;    return 0;}int main(){    int a[12][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},{0,0,0,0}};    //将每种图形储存在数组中    int m,n,i,j,k,c;    while(cin>>m>>n)    {        if(m==-1&&n==-1)      //不用连等            break;        int b[60][60];        char b1;        for(i=0; i<m; i++)            for(j=0; j<n; j++)            {                cin>>b1;                b[i][j]=b1-'A';            }        for(i=0; i<=m; i++)            b[i][n]=11;        for(j=0; j<=n; j++)            b[m][j]=11;        c=(m+1)*(n+1);        for(i=1; i<=c; i++)            fa[i]=i;        for(i=0; i<m; i++)          //比较前m行，n列        {            for(j=0; j<n; j++)            {                if(a[b[i][j]][1]&&a[b[i+1][j]][0])                    unionf(i*n+j+1,(i+1)*n+j+1);//a[i][j]的下边，a[i+1][j]的上边都为1，可合并                if(a[b[i][j]][3]&&a[b[i][j+1]][2])//a[i][j]的右边，a[i][j+1]的左边都为1，可合并                    unionf(i*n+j+1,i*n+j+2);            }        }        k=0;        for(i=1; i<=c; i++)        {            if(fa[i]==i)                k++;        }        k=k-m-n-1;        cout<<k<<endl;    }    return 0;}