Learn C++ from a young age! Record the questions in the preparation and study process for the USACO (American Informatics Olympiad) exam and record every moment.
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[Title description]
Due to recent rainfall, rainwater has collected in different places on Farmer John's field. We use a N×M(1≤< a i=5>N≤100,1≤M≤100). Each grid contains either water (W) or dry land (.). A grid is connected to the eight grids around it, and a group of connected grids is treated as a puddle. John wanted to find out how many puddles had formed in his field. Give a diagram of John's field and determine how many puddles there are.
【enter】
Input line 1: two space-separated integers: N and M.
Row 2 to N+1: Each row M< a i=4> characters, each character is W or ., they represent a row in the grid diagram. There are no spaces between characters.
【Output】
Output a line indicating the number of puddles.
【Input sample】
10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.【Output sample】
3[Detailed code explanation]
#include <bits/stdc++.h>
using namespace std;
int n, m, ans=0, vis[105][105]={0};
int dx[8]={-1,-1,-1,0,0,1,1,1}, dy[8]={-1,0,1,-1,1,-1,0,1}; // 定义8个方向的偏移坐标
char a[105][105];
struct node {
int x, y;
};
int main()
{
cin >> n >> m; // 输入n和m
for (int i=1; i<=n; i++) { // 双重for循环遍历矩阵
for (int j=1; j<=m; j++) {
cin >> a[i][j]; // 输入每个位置的字符，W或.
}
}
for (int i=1; i<=n; i++) { // 双重for循环遍历矩阵
for (int j=1; j<=m; j++) {
if (a[i][j]=='W' && vis[i][j]==0) { // 当遇到水坑'W'，且没有访问过时
ans++; // 水坑数量自增1
queue<node> q; // 定义队列
node tp = {i, j}; // 定义tp节点，此处用结构体来记录x和y坐标
q.push(tp); // 压入队列中
vis[i][j]=1; // 并标记该位置已来过
while (!q.empty()) { // 当队列不为空
node tp = q.front(); // 取出队头
q.pop();
for (int k=0; k<8; k++) { // 遍历8个方向
int xx = tp.x + dx[k]; // 得到移动后的x坐标
int yy = tp.y + dy[k]; // 得到移动后的y坐标
if (xx<1 || yy<1) continue; // 进行边界判断，超边界的继续循环
if (a[xx][yy]=='W' && vis[xx][yy]==0) { // 如果移动后仍为水坑'W'，且没有访问过
node t = {xx,yy}; // 定义t
q.push(t); // 将t压入队列中
vis[xx][yy]=1; // 并标记该位置已来过
}
}
}
}
}
}
cout << ans << endl; // 最后输出水坑数量
return 0;
}【operation result】
10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.
3