hdu 1198 Farm Irrigation

Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3643 Accepted Submission(s): 1595

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

Author

ZHENG, Lu

Source

Zhejiang University Local Contest 2005

Recommend

Ignatius.L

题目大意。有很多水管。由题目中给出。问至少需要多少水源才能保证整个田地都能灌溉到。

#include <stdio.h>int mem[10000];//用于建立并查集int msize[10000];//记录集合中水管数int mway[10000][5];//记录每块田里的水管类型int sum;//记录集合数目。即所需水源数int finder(int x)//找根节点{    int i,j,t;    i=x;    while(mem[i]!=i)        i=mem[i];    j=x;    while(j!=i)    {        t=mem[j];        mem[j]=i;        j=t;    }    return i;}void toone(int a,int b)//加入集合{    int r1,r2;    //printf("a:%d b:%d\n",a,b);    r1=finder(a);    r2=finder(b);    //printf("r1:%d r2:%d\n",r1,r2);    if(r1==r2)        return ;    if(msize[r1]>msize[r2])    {        mem[r2]=r1;        msize[r1]+=msize[r2];    }    else    {        mem[r1]=r2;        msize[r2]+=msize[r1];    }    sum--;}int main(){    int i,j,t,m,n;    char c;    while(scanf("%d%d",&m,&n)!=EOF)    {        getchar();        if(m==-1&&n==-1)            break;        sum=m*n;//sum初始为合并前的集合数        for(i=1; i<=sum; i++)        {            msize[i]=1;            mem[i]=i;            for(j=1; j<=4; j++)                mway[i][j]=0;        }        for(i=1; i<=m; i++)        {            for(j=1; j<=n; j++)            {                scanf("%c",&c);                t=(i-1)*n+j;//一维数组转化为二维数组。方便并查集。                //注意不是i*j呀！！                switch(c)//记录水管类型                {                case 'A':                    mway[t][1]=mway[t][2]=1;                    break;                case 'B':                    mway[t][1]=mway[t][3]=1;                    break;                case 'C':                    mway[t][4]=mway[t][2]=1;                    break;                case 'D':                    mway[t][4]=mway[t][3]=1;                    break;                case 'E':                    mway[t][4]=mway[t][1]=1;                    break;                case 'F':                    mway[t][3]=mway[t][2]=1;                    break;                case 'G':                    mway[t][1]=mway[t][2]=mway[t][3]=1;                    break;                case 'H':                    mway[t][1]=mway[t][2]=mway[t][4]=1;                    break;                case 'I':                    mway[t][4]=mway[t][2]=mway[t][3]=1;                    break;                case 'J':                    mway[t][1]=mway[t][4]=mway[t][3]=1;                    break;                case 'K':                    mway[t][1]=mway[t][2]=mway[t][3]=mway[t][4]=1;                }            }            getchar();        }        for(i=1; i<=m; i++)            for(j=1; j<=n; j++)            {                t=(i-1)*n+j;                if(mway[t][1]==1)//如果相通则加入集合                    if(i>1&&mway[(i-2)*n+j][4]==1)//ио                        toone(t,(i-2)*n+j);                if(mway[t][2]==1)                    if(j>1&&mway[(i-1)*n+j-1][3]==1)//вС                        toone(t,(i-1)*n+j-1);                if(mway[t][3]==1)                    if(j<n&&mway[(i-1)*n+j+1][2]==1)//ср                        toone(t,(i-1)*n+j+1);                if(mway[t][4]==1)                    if(i<m&&mway[i*n+j][1]==1)//об                        toone(t,i*n+j);            }        printf("%d\n",sum);    }    return 0;}

改良后的代码

#include <stdio.h>int mem[10000];//用于建立并查集int msize[10000];//记录集合中水管数int mway[10000][5];//记录每块田里的水管类型int sum;//记录集合数目。即所需水源数int finder(int x)//找根节点加路径压缩{    int i,j,t;    i=x;    while(mem[i]!=i)        i=mem[i];    j=x;    while(j!=i)    {        t=mem[j];        mem[j]=i;        j=t;    }    return i;}void toone(int a,int b)//加入集合{    int r1,r2;    //printf("a:%d b:%d\n",a,b);    r1=finder(a);    r2=finder(b);    //printf("r1:%d r2:%d\n",r1,r2);    if(r1==r2)        return ;    if(msize[r1]>msize[r2])    {        mem[r2]=r1;        msize[r1]+=msize[r2];    }    else    {        mem[r1]=r2;        msize[r2]+=msize[r1];    }    sum--;}int main(){    int i,j,t,m,n;    char c;    while(scanf("%d%d",&m,&n)!=EOF)    {        getchar();        if(m==-1&&n==-1)            break;        sum=m*n;//sum初始为合并前的集合数        for(i=1; i<=sum; i++)        {            msize[i]=1;            mem[i]=i;            for(j=1; j<=4; j++)                mway[i][j]=0;        }        for(i=1; i<=m; i++)        {            for(j=1; j<=n; j++)            {                scanf("%c",&c);                t=(i-1)*n+j;//一维数组转化为二维数组。方便并查集。                //注意不是i*j呀！！                //这样按方向标记代码更简洁                 if(c=='A'||c=='B'||c=='E'||c=='G'||c=='H'||c=='J'||c=='K')                    mway[t][1]=1;//上                 if(c=='A'||c=='C'||c=='F'||c=='G'||c=='H'||c=='I'||c=='K')                    mway[t][2]=1;//左                 if(c=='B'||c=='D'||c=='F'||c=='G'||c=='I'||c=='J'||c=='K')                    mway[t][3]=1;//右                 if(c=='C'||c=='D'||c=='E'||c=='H'||c=='I'||c=='J'||c=='K')                    mway[t][4]=1;//下            }            getchar();        }        for(i=1; i<=m; i++)            for(j=1; j<=n; j++)            {                t=(i-1)*n+j;                if(mway[t][1]==1)//如果相通则加入集合                    if(i>1&&mway[(i-2)*n+j][4]==1)//查找了上就不用查找下了                        toone(t,(i-2)*n+j);                if(mway[t][2]==1)                    if(j>1&&mway[(i-1)*n+j-1][3]==1)//查找了左就不用查找右了                        toone(t,(i-1)*n+j-1);            }        printf("%d\n",sum);    }    return 0;}