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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
#270459	#7733. Cool, It’s Yesterday Four Times More	March_April	WA	211ms	18936kb	C++20	3.1kb	2023-11-30 21:25:41	2023-11-30 21:25:41

Judging History

你现在查看的是最新测评结果

[2023-11-30 21:25:41]
评测

测评结果：WA
用时：211ms
内存：18936kb

查看

[2023-11-30 21:25:41]
提交

answer

#include <bits/stdc++.h>

using namespace std;

struct Node
{
    int ans[4];
};

void solve()
{
    int n, m;
    cin >> n >> m;
    vector<string> g(n);

    for(int i = 0;i < n; i ++)      cin >> g[i];

    int dx[] = {0, 0, 1, -1}, dy[] = {1, -1, 0, 0};

    vector<vector<int>> block(n, vector<int> (m));//每个格子属于哪个连通块
    int cnt = 0;//连通块的数量
    
    vector<int> idx(1010);
    auto bfs = [&] (int x, int y, int id)
    {
        queue<pair<int, int>> q;
        q.push({x, y});
        block[x][y] = id;

        while(q.size())
        {
            auto [ix, iy] = q.front();
            q.pop();
            idx[id] ++;

            for(int i = 0;i < 4; i ++)
            {
                int a = dx[i] + ix, b = dy[i] + iy;
                if(a < 0 || a >= n || b < 0 || b >= m)        
                    continue;
                
                if(block[a][b] || g[a][b] == 'O')     continue;

                q.push({a, b});
                block[a][b] = id;
            }
        }
    };
    
    for(int i = 0;i < n; i ++)
        for(int j = 0;j < m; j ++)
            if(!block[i][j] && g[i][j] != 'O')
                bfs(i, j, ++ cnt);

    vector<pair<int, int>> S[cnt + 1];

    for(int i = 0;i < n; i ++)
        for(int j = 0;j < m; j ++)
            S[block[i][j]].push_back({i, j});


    // for(int i = 1;i <= cnt; i ++)
    // {
    //     //clog << S[i].size() << '\n';
    //     for(auto [x, y] : S[i])
    //         clog << x << ' ' << y << ' ' << i << '\n';
    // }

    int ans = 0;

    auto check = [&] (vector<pair<int, int>> A, vector<pair<int, int>> B, int ddx, int ddy)
    {
        vector<vector<int>> st(n + 2000, vector<int> (m + 2000));
        for(auto [x, y] : B)
            st[x + ddx + 1000][y + ddy + 1000] = true;
            
        for(auto [x, y] : A)
            if(st[x][y])
                return false;
        
        return B.size() >= A.size();
    };

    for(int i = 1;i <= cnt; i ++)
    {
        bool M = true;
        for(int j = 1;j <= cnt; j ++)
        {
            if(S[j].size() < S[i].size() || i == j)   
                continue;
            bool ok = true;

            for(auto [x, y] : S[i])//这个和j的左上角对齐
            {
                for(auto [xx, yy] : S[j])
                {
                    if(check(S[i], S[j], -x + xx, -y + yy))
                    {
                        ok = false;
                        break;
                    }
                    if(!ok)
                        break;
                }
            }
            if(!ok) 
            {
                M = false;
                break;
            }
        }
        if(M)  
        {
            //clog << idx[i] << ' ' << i << '\n';
            ans += idx[i];
        }
    }

    cout << ans << '\n';

}           

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

    int tt;
    cin >> tt;
    while(tt --)
    {
        solve();
    }
    return 0;
}

源文件, 语言: C++20

详细

Test #1:

score: 100

Accepted

time: 19ms

memory: 18900kb

input:

4
2 5
.OO..
O..O.
1 3
O.O
1 3
.O.
2 3
OOO
OOO

output:

result:

ok 4 lines

Test #2:

score: -100

Wrong Answer

time: 211ms

memory: 18936kb

input:

200
2 4
OOO.
OO..
2 3
OOO
.O.
3 3
O.O
OOO
OO.
4 1
.
.
O
O
1 2
.O
1 1
.
2 5
.OO..
.O.O.
2 1
O
O
1 1
O
1 3
.OO
5 1
O
O
.
O
.
5 2
O.
..
O.
.O
..
5 3
...
...
.OO
..O
OOO
3 5
..O.O
.O.O.
.OO.O
5 2
.O
OO
O.
O.
..
2 1
O
O
3 5
.O.OO
O...O
..OO.
1 5
.....
5 1
O
.
O
.
.
5 3
OOO
OO.
.OO
OO.
O.O
2 1
O
.
5 2
O.
...

output:

result:

wrong answer 12th lines differ - expected: '7', found: '4'