Ahws_rwhy @ 2024-10-26 20:27:58
#include <bits/stdc++.h>
using namespace std;
#define int long long
int t;
int n, m, k;
int x, y, d;
char mp[1001][1001];
int flag[1001][1001];
int ans;
signed main() {
cin >> t;
while (t--) {
memset(flag,0,sizeof flag);
ans = 0;
cin >> n >> m >> k;
cin >> x >> y >> d;
int nowx = x, nowy = y;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
cin >> mp[i][j];
}
}
while (k > 0) {
if (d == 0) {
k--;
nowx = nowx, nowy = nowy + 1;
if (nowx > n || nowy < 1 || nowx < 1 || nowy > m) {
d = (d + 1) % 4;
nowy = nowy - 1;
// continue;
}
if (mp[nowx][nowy] == '.' && d == 0 && flag[nowx][nowy] == 0) {
ans++;
flag[nowx][nowy] = 1;
}
if (mp[nowx][nowy] == 'x' && d == 0) {
d = (d + 1) % 4;
nowy = nowy - 1;
}
}
if (d == 1) {
k--;
nowx = nowx + 1, nowy = nowy;
if (nowx > n || nowy < 1 || nowx < 1 || nowy > m) {
d = (d + 1) % 4;
nowx = nowx - 1;
// continue;
}
if (mp[nowx][nowy] == '.' && d == 1 && flag[nowx][nowy] == 0) {
ans++;
flag[nowx][nowy] = 1;
}
if (mp[nowx][nowy] == 'x' && d == 1) {
d = (d + 1) % 4;
nowx = nowx - 1;
}
}
if (d == 2) {
k--;
nowx = nowx, nowy = nowy - 1;
if (nowx > n || nowy < 1 || nowx < 1 || nowy > m) {
d = (d + 1) % 4;
nowy = nowy + 1;
// continue;
}
if (mp[nowx][nowy] == '.' && d == 2 && flag[nowx][nowy] == 0) {
ans++;
flag[nowx][nowy] = 1;
}
if (mp[nowx][nowy] == 'x' && d == 2) {
d = (d + 1) % 4;
nowy = nowy + 1;
}
}
if (d == 3) {
k--;
nowx = nowx - 1, nowy = nowy;
if (nowx > n || nowy < 1 || nowx < 1 || nowy > m) {
d = (d + 1) % 4;
nowx = nowx + 1;
// continue;
}
if (mp[nowx][nowy] == '.' && d == 3 && flag[nowx][nowy] == 0) {
ans++;
flag[nowx][nowy] = 1;
}
if (mp[nowx][nowy] == 'x' && d == 3) {
// k--;
d = (d + 1) % 4;
nowx = nowx + 1;
}
}
}
if(1 <= x && x <= n && 1 <= y && y <= m) ans++;
cout << ans << endl;
}
return 0;
}by hyl2718281828 @ 2024-10-26 20:34:12
感觉可能是ans算重了，但不确定
by hyl2718281828 @ 2024-10-26 20:35:21
我建议你定义一个常量数组，专门存横纵坐标变化量，直接一个循环就可以跑四种情况
by Ahws_rwhy @ 2024-10-26 20:35:48
@hyl2718281828 能讲讲那算重了吗
by hyl2718281828 @ 2024-10-26 20:36:20
还有就是ans没必要一个个加，容易出错，你可以只记录走过的点，最后遍历一遍flag数组加上去，绝对不会错
by hyl2718281828 @ 2024-10-26 20:37:41
另外，对于不可以走的点，我觉得在走到那之前就应该把它排除。可以专门写一个check函数判断是否可行，代码会简洁很多
by hyl2718281828 @ 2024-10-26 20:39:42
还有他转向是单独算一步的，你是不是没算进去
AC代码：
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3+1;
const int dx[] = {0,1,0,-1}, dy[] = {1,0,-1,0};
int T,n,m,k,ans;
char G[N][N];
bool vis[N][N];
bool check(int x,int y) {return x > 0 && x <= n && y > 0 && y <= m && G[x][y] == '.';}
int main() {
freopen("explore.out","w",stdout);
scanf("%d",&T);
while (T--) {
int x,y,d;
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 >> G[i][j];
vis[x][y] = 1;
for (int r = 1; r <= k; r++) {
int _x = x+dx[d], _y = y+dy[d];
if (check(_x,_y)) {
vis[_x][_y] = 1;
x = _x;
y = _y;
} else d = (d+1)%4;
}
for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++) ans += vis[i][j];
printf("%d\n",ans);
ans = 0;
memset(vis,0,sizeof(vis));
}
return 0;
}by hyl2718281828 @ 2024-10-26 20:40:22
多打了一个freopen
by Ahws_rwhy @ 2024-10-26 20:41:19
@hyl2718281828 我觉得算进去了呀，每次if 里面 的次数都减 1 了呀？
by hyl2718281828 @ 2024-10-26 20:45:50
哦我好像知道了，你走到障碍之后再退回来多算了一步，因为只有k步，你这里多占了一步，没加回去，后面ans就少了
by Ahws_rwhy @ 2024-10-26 20:49:13
@hyl2718281828 现在过了，谢谢（猫的，出发点未打标记）