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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
#270323	#7733. Cool, It’s Yesterday Four Times More	March_April^#	WA	1ms	3816kb	C++20	2.8kb	2023-11-30 18:50:03	2023-11-30 18:50:03

Judging History

你现在查看的是最新测评结果

[2023-11-30 18:50:03]
评测

测评结果：WA
用时：1ms
内存：3816kb

查看

[2023-11-30 18:50:03]
提交

answer

#include <bits/stdc++.h>

using namespace std;

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>> p(cnt + 1);

    auto bfs1 = [&] (int x, int y, int d)
    {
        queue<pair<int, int>> q;

        q.push({x, y});

        int val = 0;
        while(q.size())
        {
            auto [ix, iy] = q.front();
            q.pop();

            val ++;

            int a = dx[d] + ix, b = dy[d] + iy;

            // if(ix == 1 && iy == 4 && d == 3)
            //     clog << a << ' ' << b << '\n';

            if(a < 0 || a >= n || b < 0 || b >= m)        
                    continue;
            if(g[a][b] == 'O')     continue;

            // if(ix == 1 && iy == 4 && d == 3)
            //     clog << a << ' ' << b << '\n';

            q.push({a, b});
        }

        return val;
    };

    for(int i = 0;i < n; i ++)
        for(int j = 0;j < m; j ++)
        {
            if(g[i][j] == 'O')  continue;
            p[block[i][j]].first = max(bfs1(i, j, 0), p[block[i][j]].first);
            p[block[i][j]].second = max(bfs1(i, j, 3), p[block[i][j]].second);
        }
    
    int ans = 0;
    for(int i = 1;i <= cnt; i ++)
    {
        //clog << p[i].first << ' ' << p[i].second << '\n';
        bool ok = true;   
        for(int j = 1;j <= cnt; j ++)
        {
            if(i != j)
                if(p[j].first >= p[i].first && p[j].second >= p[i].second)
                    ok = false;
        }
        if(ok)  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: 0ms

memory: 3816kb

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: 1ms

memory: 3596kb

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'