ZOJ2412 Farm Irrigation (DFS)

xiaoxiao2021-02-28  2

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


-1 -1

Sample Output 2 3

题意:求得最少水源数 能使所有格子都被灌溉


#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; char mp[55][55]; int n,m,book[55][55],nx[4][2]={0,1,-1,0,0,-1,1,0};//注意顺序 int zt[][4]={{0,1,1,0},{1,1,0,0},{0,0,1,1},{1,0,0,1}, {0,1,0,1},{1,0,1,0},{1,1,1,0},{0,1,1,1}, {1,0,1,1},{1,1,0,1},{1,1,1,1}};//将右设为起点,逆时针数 zt(状态)数组保存这十一种不同状态的格子里的水管(1表示有开口0则是没有) void dfs(int x,int y,int k) { book[x][y]=1; for(int i=0;i<4;i++) { if(zt[k][i]) { int tx=x+nx[i][0]; int ty=y+nx[i][1]; int tt=mp[tx][ty]-'A';//tt表示下一个格子。注意对应zt数组 if(tx>=0&&ty>=0&&tx<n&&ty<m&&book[tx][ty]==0&&zt[tt][(i+2)%4])//zt[v][(2+i)%4]判断两个格子里的进出口是否连通 dfs(tx,ty,tt); } } } int main() { while(~scanf("%d%d",&n,&m)&&n!=-1&&m!=-1) { memset(book,0,sizeof(book)); for(int i=0;i<n;i++) scanf("%s",mp[i]); int sum=0; for(int i=0;i<n;i++) { for(int j=0;j<m;j++) { if(book[i][j]==0) { int tt=mp[i][j]-'A'; dfs(i,j,tt); sum++; } } } printf("%d\n",sum); } return 0; }


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