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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
#243113	#6723. Grid with Arrows	Cipherxzc^#	WA	72ms	15016kb	C++20	2.7kb	2023-11-07 21:10:19	2023-11-07 21:10:19

Judging History

你现在查看的是最新测评结果

[2023-11-07 21:10:19]
评测

测评结果：WA
用时：72ms
内存：15016kb

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[2023-11-07 21:10:19]
提交

answer

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

const int N = 1e5 + 5, dx[] = {-1, 1, 0, 0}, dy[] = {0, 0, -1, 1};
int T, n, m, head[N], nxt[N * 2], to[N * 2], tot;
bool vis[N];
vector<vector<int>> a, dir;
vector<vector<bool>> in;

inline void add(int u, int v){
    //cout << u << ' ' << v << endl;
    to[++tot] = v;
    nxt[tot] = head[u];
    head[u] = tot;
}

void dfs(int p){
    //cout << p << ' ';
    vis[p] = true;
    tot++;
    for (int i = head[p]; i; i = nxt[i]){
        int q = to[i];
        if (!vis[q]){
            dfs(q);
        }
    }
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);

    cin >> T;
    while (T--){
        cin >> n >> m;
        a.resize(n + 1);
        dir.resize(n + 1);
        in.resize(n + 1);
        for (int i = 1; i <= n; i++){
            a[i].resize(m + 1);
            dir[i].resize(m + 1);
            in[i].resize(m + 1);
        }
        for (int i = 1; i <= n * m; i++){
            vis[i] = false;
        }

        char ch;
        for (int i = 1; i <= n; i++){
            for (int j = 1; j <= m; j++){
                cin >> ch;
                if (ch == 'u'){
                    dir[i][j] = 0;
                }else if (ch == 'd'){
                    dir[i][j] = 1;
                }else if (ch == 'l'){
                    dir[i][j] = 2;
                }else{
                    dir[i][j] = 3;
                }
            }
        }
        for (int i = 1; i <= n; i++){
            for (int j = 1; j <= m; j++){
                cin >> a[i][j];
            }
        }

        tot = 0;
        int cnt = 0;
        for (int i = 1; i <= n; i++){
            for (int j = 1; j <= m; j++){
                int gg = dir[i][j], x = i + dx[gg] * a[i][j], y = j + dy[gg] * a[i][j];
                //cout << x << ' ' << y << endl;
                if (1 <= x && x <= n && 1 <= y && y <= m){
                    if (!in[x][y]){
                        cnt++;
                        in[x][y] = true;
                    }
                    add((i - 1) * m + j, (x - 1) * m + y);
                }
            }
        }

        tot = 0;
        if (cnt < n * m - 1){
            cout << "No" << endl;
            continue;
        }else if (cnt < n * m){
            for (int i = 1; i <= n; i++){
                for (int j = 1; j <= m; j++){
                    if (!in[i][j]){
                        dfs((i - 1) * m + j);
                    }
                }
            }
        }else{
            dfs(1);
        }

        if (tot == n * m){
            cout << "Yes" << endl;
        }else{
            cout << "No" << endl;
        }
    }

    return 0;
}

源文件, 语言: C++20

詳細信息

Test #1:

score: 100

Accepted

time: 0ms

memory: 3448kb

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

Wrong Answer

time: 72ms

memory: 15016kb

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
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No...

result:

wrong answer expected YES, found NO [5th token]