56 lines
1.2 KiB
C++
56 lines
1.2 KiB
C++
#include <bits/stdc++.h>
|
||
|
||
/**
|
||
3 4
|
||
P P P P
|
||
P O P X
|
||
P O X P
|
||
O P P P
|
||
|
||
答案：10
|
||
*/
|
||
using namespace std;
|
||
const int N = 110;
|
||
char a[N][N];
|
||
struct Node {
|
||
int x;
|
||
int y;
|
||
int step;
|
||
};
|
||
queue<Node> q;
|
||
int dx[] = {0, 0, -1, 1};
|
||
int dy[] = {-1, 1, 0, 0};
|
||
int cnt;
|
||
|
||
int main() {
|
||
int n, m;
|
||
cin >> n >> m;
|
||
for (int i = 1; i <= m; i++)
|
||
for (int j = 1; j <= m; j++) {
|
||
cin >> a[i][j];
|
||
//记录初始病人
|
||
if (a[i][j] == 'X') q.push({i, j, 0});
|
||
}
|
||
//n分钟
|
||
while (!q.empty()) {
|
||
auto t = q.front();
|
||
q.pop();
|
||
//执行n分钟
|
||
if (t.step == n) break;
|
||
//四个方向
|
||
for (int i = 0; i <= 3; i++) {
|
||
int tx = t.x + dx[i];
|
||
int ty = t.y + dy[i];
|
||
if (tx >= 1 && tx <= m && ty >= 1
|
||
&& ty <= m && a[tx][ty] == 'P') {
|
||
a[tx][ty] = 'X';
|
||
q.push({tx, ty, t.step + 1});
|
||
}
|
||
}
|
||
}
|
||
for (int i = 1; i <= m; i++)
|
||
for (int j = 1; j <= m; j++)
|
||
if (a[i][j] == 'X')cnt++;
|
||
printf("%d ", cnt);
|
||
return 0;
|
||
} |