Number of NYOJ27 pools

Number of pools

Time limit: ms | Memory limit:65535 KB

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

At first, I didn't think to modify the values on the map, but instead used an array, which is more complex than the standard path.

#include  <iostream>using namespace std;int m1,m2;int dp[100][100],map[100][100] ;//,DIRECT[4][2];INT&NBSP;DIRECT[4][2]={{1,0},{0,1},{-1,0},{0,-1}};VOID&NBSP;DP (int x, int y) { Int k,xx,yy;dp[x][y]=1;for (k=0;k<4;k++) {xx=x+direct[k][0];yy=y+direct[k][1];if (xx>=0&&xx <m1 && yy>=0&&yy<m2 && 1==map[xx][yy] && &NBSP;0==DP[XX][YY]) DP (XX,YY);}} Int search (int x, int y) {int k,xx,yy;if (1==map[x][y])  &&  (0==dp[x] [y])) {dp[x][y]=1;for (k=0;k<4;k++) {xx=x+direct[k][0];yy=y+direct[k][1];//yy>=0&&yy<m2  was written  yy>0&&yy<m2  wasted more than 2 hours .... if (xx>=0&&xx<m1 && yy>=0&&yy<m2 && 1==map[xx][ YY]&NBSP;&AMP;&AMP;&NBSP;0==DP[XX][YY]) DP (XX,YY);} return 1;} return 0;} Int main () {int n,i,j,count;cin>>n;while (n--) {//Build map cin>>m1>>m2;for (i=0;i<m1;i++) for (j=0;j<m2;j++) {cin>>map[i][j];dp [i][j]=0;} Find the Pool count=0;for (i=0;i<m1;i++) for (j=0;j<m2;j++) if (Search (i,j)) ++count;cout<<count<<endl;} return 0;}

After I've seen the label, I changed it.

#include <iostream>using namespace Std;int map[102][102];int direct[4][2]={{1,0},{0,1},{-1,0},{0,-1}};int M1, M2;void Search (int x, int y) {int k,xx,yy;for (k=0;k<4;k++) {xx=x+direct[k][0];yy=y+direct[k][1];if (xx>=0& &XX<M1 && yy>=0&&yy<m2 && 1==map[xx][yy]) {map[xx][yy]=0; Search (XX,YY);}} int main () {int N,i,j,count;cin>>n;while (n--) {//Build map Cin>>m1>>m2;for (i=0;i<m1;i++) for (j=0;j <m2;j++) cin>>map[i][j];count=0;//Find pool for (i=0;i<m1;i++) for (j=0;j<m2;j++) {if (Map[i][j]) {count++; map[i][j]=0; Search (I,J);}} Cout<<count<<endl;} return 0;}

This article is from the "hacker" blog, make sure to keep this source http://anglecode.blog.51cto.com/5628271/1631173

Number of NYOJ27 pools