模拟
第一眼,可能有人回想起 dfs.
但因为起点终点,并且走的步数都告诉你了,所以直接模拟就行.
注意
起始点也算被走过,所以可以用一个标记数组,判断当前格子有没有被走过.
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dir[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}}; // 方向,x,y
int T;
char a[1003][1003];
int n;
int m;
int k;
int d;
int x, y;
int main() {
scanf("%d", &T);
while (T--) {
scanf("%d%d%d", &n, &m, &k);
scanf("%d%d%d", &x, &y, &d);
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++)
cin >> a[i][j];
}
bool vis[1003][1003];
memset(vis, 0, sizeof(vis));
int ans = 1; // 起始点也算被走过
vis[x][y] = 1;
for (int i = 1; i <= k; i++) {
int nx = x + dir[d][0];
int ny = y + dir[d][1];
// 判断是否合法
if (1 <= nx && nx <= n && 1 <= ny && ny <= m && a[nx][ny] != 'x') {
if (!vis[nx][ny]) {
ans++;
vis[nx][ny] = 1;
}
x = nx, y = ny; // 更新坐标
}
else
d = (d + 1) % 4; // 更改方向
}
printf("%d\n", ans);
}
return 0;
}
标签:P11228,int,题解,scanf,d%,2024,vis,include,1003
From: https://www.cnblogs.com/panda-lyl/p/18519912