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QOJ

ID	题目	提交者	结果	用时	内存	语言	文件大小	提交时间	测评时间
#499463	#6723. Grid with Arrows	Umok	RE	1ms	8064kb	C++20	1.5kb	2024-07-31 14:31:59	2024-07-31 14:31:59

Judging History

你现在查看的是最新测评结果

[2024-07-31 14:31:59]
评测

测评结果：RE
用时：1ms
内存：8064kb

查看

[2024-07-31 14:31:59]
提交

answer

#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
const int N = 1e4 + 5;
#define int long long
typedef pair<int, int> PII;
#define MAX LONG_LONG_MAX
string mp[N];
int ar[N][N], f[N][N];
bool st[N][N];
int n, m;

int dfs(int x, int y)
{
    if (x > n || x < 1 || y > m || y < 1)
        return 0;
    if (f[x][y])
        return f[x][y] + 1;
    if (st[x][y])
        return 1;
    st[x][y] = 1;
    int xx = x, yy = y;
    if (mp[x][y] == 'r')
        yy += ar[x][y];
    else if (mp[x][y] == 'd')
        xx += ar[x][y];
    else if (mp[x][y] == 'l')
        yy -= ar[x][y];
    else
        xx -= ar[x][y];

    f[x][y] = dfs(xx, yy) + 1;
    st[x][y] = 0;
    return f[x][y];
}

void solve()
{
    cin >> n >> m;

    for (int i = 1; i <= n; i++)
    {
        cin >> mp[i];
        mp[i] = " " + mp[i];
    }
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            scanf("%lld", &ar[i][j]), f[i][j] = 0;

    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
        {
            if (f[i][j])
                continue;
            dfs(i, j);
        }

    int ans = m * n;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            if (f[i][j] == ans)
            {
                cout << "Yes" << endl;
                return;
            }
    cout << "No" << endl;
}
signed main()
{
    int tcase;
    cin >> tcase;
    while (tcase--)
        solve();
    return 0;
}

源文件, 语言: C++20

详细

Test #1:

score: 100

Accepted

time: 1ms

memory: 8064kb

input:

2
2 3
rdd
url
2 1 1
1 1 2
2 2
rr
rr
1 1
1 1

output:

Yes
No

result:

ok 2 token(s): yes count is 1, no count is 1

Test #2:

score: -100

Runtime Error

input:

1109
5 8
rddddldl
drruludl
rrldrurd
urrrlluu
uurrulrl
4 4 1 2 4 3 1 6
1 3 5 1 1 1 3 6
2 4 1 1 2 1 1 1
2 3 4 2 4 3 3 3
4 1 1 2 2 5 1 5
7 9
rdrddrdll
urrdruldl
ruullrulu
drrlrlddl
rrrdddlll
ruulururl
ruurrlluu
7 1 1 1 2 1 2 3 6
1 7 3 1 3 1 2 1 8
2 2 1 2 4 3 1 2 2
2 2 4 1 1 5 3 3 1
3 4 6 1 2 1 2 7 7
6 ...

output:

Yes
No
Yes
No
Yes
No
No
Yes
Yes
Yes
No
No
No
Yes
Yes
No
Yes
No
Yes
No
No
No
Yes
Yes
No
Yes
Yes
No
No
No
No
Yes
No
No
Yes
No
No
No
Yes
Yes
Yes
Yes
No
Yes
No
No
No
Yes
Yes
No
Yes
No
Yes
No
Yes
No
No
No
Yes
No
Yes
Yes
Yes
Yes
No
Yes
Yes
No
Yes
Yes
No
No
No
Yes
No
No
Yes
Yes
No
No
Yes
Yes
Yes
Yes
Yes
No...