zoj2412 & hdu1198 - Farm Irrigation (DFS实现)

Farm Irrigation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6029    Accepted Submission(s): 2608

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

                  这道题可以用并查集来做，下面的代码是用dfs实现的，比较尴尬的一点是画图有点费事，恶心死我了。

/***********************************   acm:   zoj-2412 & hdu-1198**   title: Farm Irrigation**   time : 2014.11.18**********************************///考察深度优先搜索#include <stdio.h>#include <stdlib.h>#define MAXVEX 51int map[MAXVEX*3][MAXVEX*3];// 1表示有路，0表示无路/*  将数组合并比较好，否则还要有个switch判断 比较麻烦int A[3][3] ={{0,1,0},{1,1,0},{0,0,0}};int B[3][3] ={{0,1,0},{0,1,1},{0,0,0}};int C[3][3] ={{0,0,0},{1,1,0},{0,1,0}};int D[3][3] ={{0,0,0},{0,1,1},{0,1,0}};int E[3][3] ={{0,1,0},{0,1,0},{0,1,0}};int F[3][3] ={{0,0,0},{1,1,1},{0,0,0}};int G[3][3] ={{0,1,0},{1,1,1},{0,0,0}};int H[3][3] ={{0,1,0},{1,1,0},{0,1,0}};int I[3][3] ={{0,0,0},{1,1,1},{0,1,0}};int J[3][3] ={{0,1,0},{0,1,1},{0,1,0}};int K[3][3] ={{0,1,0}, {1,1,1},{0,1,0}};*/int square[11][3][3]={{{0,1,0},{1,1,0},{0,0,0}},{{0,1,0},{0,1,1},{0,0,0}},{{0,0,0},{1,1,0},{0,1,0}},{{0,0,0},{0,1,1},{0,1,0}},{{0,1,0},{0,1,0},{0,1,0}},{{0,0,0},{1,1,1},{0,0,0}},{{0,1,0},{1,1,1},{0,0,0}},{{0,1,0},{1,1,0},{0,1,0}},{{0,0,0},{1,1,1},{0,1,0}},{{0,1,0},{0,1,1},{0,1,0}},{{0,1,0}, {1,1,1},{0,1,0}}};int dir[4][2] ={{0,1},{0,-1},{1,0},{-1,0}};   //4个方向int M, N;  //长 宽void DFS(int x, int y){    int i;    int xx, yy;    for (i = 0; i < 4; i++)    {        xx = x + dir[i][0];        yy = y + dir[i][1];        if (xx < 0 || xx >= M*3 || yy < 0 || yy >= N*3)        {            continue;        }        if (map[xx][yy] == 1)        {            map[xx][yy] = 0;            DFS(xx, yy);        }    }}int main(){    int i, j;    int k, z;    char c;    int count;    while (scanf("%d%d", &M, &N), M > 0 && N > 0)    {        for (i = 0; i < M; i++)   //画地图        {            for (j = 0; j < N; j++)            {                scanf(" %c", &c);                for (k =0; k < 3; k++)                {                    for (z = 0; z < 3; z++)                    {                        map[k + i*3][z + j*3] = square[c-'A'][k][z];                    }                }            }        }        count = 0;        for (i = 0; i < M*3; i++)        {            for (j = 0; j < N*3; j++)            {                if (map[i][j] == 1)                {                    map[i][j] = 0;                    DFS(i, j);                    count++;                }            }        }        printf("%d\n", count);    }    return 0;}