Number of nyoj27 pools (DFS)

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Number of sinks time limit: theMs | Memory Limit:65535KB Difficulty:4

Describe

Nanyang Polytechnic Campus There are some rivers and some lakes, now, we think of them as a pool, suppose there is a map of our school, this map only to identify whether this is a pool, now, your task came, please use the computer to calculate the total number of pools in the map.

Input

the first line enters an integer n, representing a total of n sets of test data
Each group of data is entered first the number of rows of the map m (0<m<100) and the number of columns N (0<n<100), and then enter the next m row each row input n number, indicating whether there is water or no water (1 means this is the pool, 0 means the ground)

Output

outputs the number of pools in the map.
It is important to note that each pool side (up and down four positions) can be considered to be the same pool if it is a sink.

Sample input

23 41 0 0 0 0 0 1 11 1 1 05 51 1 1 1 00 0 1 0 10 0 0 0 01 1 1 0 00 0 1 1 1

Sample output

23

Source

[Zhang Yunzun] Original

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Zhang Yunzun

Go through the map first to find the pool and deep search, the pool next to it all marked, at the same time the number of pools +1

Very simple. But remember this is about four directions up and down. Not eight directions 0.0

#include <stdio.h> #include <string.h>int t,m,n,i,j,count;int visit[100][100],a[100][100];void dfs (int s1 , int s2) {if (s1<0| | s1==m| | s2<0| | s2==n| | visit[s1][s2]| |! A[S1][S2]) return, Visit[s1][s2]=1;dfs (S1+1,S2);d FS (S1-1,S2);d FS (s1,s2-1);d FS (s1,s2+1);} int main () {scanf ("%d", &t), while (t--) {scanf ("%d%d", &m,&n), memset (A,0,sizeof (a)); Memset (visit,0, sizeof (visit)); for (i=0;i<m;i++) for (j=0;j<n;j++) scanf ("%d", &a[i][j]); Count=0;for (i=0;i<m;i++) for (j =0;j<n;j++) {if (a[i][j]==1&&visit[i][j]==0) Count++,dfs (i,j);} printf ("%d\n", Count);}}        

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Number of nyoj27 pools (DFS)