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 1Benny 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 2Several 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.
InputThere 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 3Author: ZHENG, Lu
好好反思这个代码为什么错误??????
Problem : 1198 ( Farm Irrigation ) Judge Status : Wrong Answer RunId : 1357129 Language : C ++ Author : forever4444 Code Render Status : Rendered By HDOJ C ++ Code Rander Version 0.01 Beta // 1863626 2009-05-10 21:31:51 Wrong Answer 2412 C++ 0 200 Wpl #include < iostream > #define MAX 55 using namespace std; int n,m,un[MAX * MAX]; char map[MAX][MAX]; int dir[ 2 ][ 2 ] = {{ 0 , - 1 },{ - 1 , 0 }}; void Init() { int i,j; for (i = 0 ;i < m;i ++ ) for (j = 0 ;j < n;j ++ ) { un[i * m + j] = i * m + j; cin >> map[i][j]; } } bool Bound( int x, int y) { if (x >= 0 && y >= 0 && x < m && y < n) return true ; else return false ; } int Find( int x) { return un[x]; } void Merge( int a, int b) { int mmax,mmin,i,temp; mmax = Find(a); mmin = Find(b); if (mmax == mmin) return ; if (mmin > mmax) { temp = mmin; mmin = mmax; mmax = temp; } for (i = 0 ;i < n * m;i ++ ) { if (un[i] == mmax) un[i] = mmin; } } int main() { int i,j,k; while (scanf( " %d%d " , & m, & n) != EOF) { if (m < 0 || n < 0 ) break ; Init(); int ti,tj,a,b; for (i = 0 ;i < m;i ++ ) for (j = 0 ;j < n;j ++ ) { for (k = 0 ;k < 2 ;k ++ ) { ti = i + dir[k][ 0 ]; tj = j + dir[k][ 1 ]; a = i * m + j; // 我用行存储 b = ti * m + tj; if ( ! Bound(ti,tj)) continue ; if (k == 0 ) // 向左 { if (map[i][j] == ' A ' || map[i][j] == ' C ' || map[i][j] == ' F ' || map[i][j] == ' G ' || map[i][j] == ' H ' || map[i][j] == ' I ' || map[i][j] == ' K ' ) { if (map[ti][tj] == ' B ' || map[ti][tj] == ' D ' || map[ti][tj] == ' F ' || map[ti][tj] == ' G ' || map[ti][tj] == ' I ' || map[ti][tj] == ' J ' || map[ti][tj] == ' K ' ) { Merge(a,b); } } } else // 向上 { if (map[i][j] == ' A ' || map[i][j] == ' B ' || map[i][j] == ' E ' || map[i][j] == ' G ' || map[i][j] == ' H ' || map[i][j] == ' J ' || map[i][j] == ' K ' ) { if (map[ti][tj] == ' C ' || map[ti][tj] == ' D ' || map[ti][tj] == ' E ' || map[ti][tj] == ' H ' || map[ti][tj] == ' I ' || map[ti][tj] == ' J ' || map[ti][tj] == ' K ' ) { Merge(a,b); } } } } } int sum = 0 ; for (i = 0 ;i < m;i ++ ) for (j = 0 ;j < n;j ++ ) { if (un[i * m + j] == (i * m + j)) sum ++ ; } printf( " %d\n " ,sum); } return 0 ; } // 1863626 2009-05-10 21:31:51 Wrong Answer 2412 C++ 0 200 Wpl #include < iostream > #define MAX 55 using namespace std; int n,m,un[MAX * MAX]; char map[MAX][MAX]; int dir[ 2 ][ 2 ] = {{ 0 , - 1 },{ - 1 , 0 }}; void Init() { int i,j; for (i = 0 ;i < m;i ++ ) for (j = 0 ;j < n;j ++ ) { un[i * m + j] = i * m + j; cin >> map[i][j]; } } bool Bound( int x, int y) { if (x >= 0 && y >= 0 && x < m && y < n) return true ; else return false ; } int Find( int x) { return un[x]; } void Merge( int a, int b) { int mmax,mmin,i,temp; mmax = Find(a); mmin = Find(b); if (mmax == mmin) return ; if (mmin > mmax) { temp = mmin; mmin = mmax; mmax = temp; } for (i = 0 ;i < n * m;i ++ ) { if (un[i] == mmax) un[i] = mmin; } } int main() { int i,j,k; while (scanf( " %d%d " , & m, & n) != EOF) { if (m < 0 || n < 0 ) break ; Init(); int ti,tj,a,b; for (i = 0 ;i < m;i ++ ) for (j = 0 ;j < n;j ++ ) { for (k = 0 ;k < 2 ;k ++ ) { ti = i + dir[k][ 0 ]; tj = j + dir[k][ 1 ]; a = i * m + j; // 我用行存储 b = ti * m + tj; if ( ! Bound(ti,tj)) continue ; if (k == 0 ) // 向左 { if (map[i][j] == ' A ' || map[i][j] == ' C ' || map[i][j] == ' F ' || map[i][j] == ' G ' || map[i][j] == ' H ' || map[i][j] == ' I ' || map[i][j] == ' K ' ) { if (map[ti][tj] == ' B ' || map[ti][tj] == ' D ' || map[ti][tj] == ' F ' || map[ti][tj] == ' G ' || map[ti][tj] == ' I ' || map[ti][tj] == ' J ' || map[ti][tj] == ' K ' ) { Merge(a,b); } } } else // 向上 { if (map[i][j] == ' A ' || map[i][j] == ' B ' || map[i][j] == ' E ' || map[i][j] == ' G ' || map[i][j] == ' H ' || map[i][j] == ' J ' || map[i][j] == ' K ' ) { if (map[ti][tj] == ' C ' || map[ti][tj] == ' D ' || map[ti][tj] == ' E ' || map[ti][tj] == ' H ' || map[ti][tj] == ' I ' || map[ti][tj] == ' J ' || map[ti][tj] == ' K ' ) { Merge(a,b); } } } } } int sum = 0 ; for (i = 0 ;i < m;i ++ ) for (j = 0 ;j < n;j ++ ) { if (un[i * m + j] == (i * m + j)) sum ++ ; } printf( " %d\n " ,sum); } return 0 ; }
转载于:https://www.cnblogs.com/forever4444/archive/2009/05/11/1454349.html
