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QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#149440#4432. Jungle Trailrgnerdplayer#WA 291ms103720kbC++202.7kb2023-08-24 16:50:442023-08-24 16:50:45

Judging History

你现在查看的是最新测评结果

  • [2023-08-24 16:50:45]
  • 评测
  • 测评结果:WA
  • 用时:291ms
  • 内存:103720kb
  • [2023-08-24 16:50:44]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;

using i64 = long long;

int main() {
    cin.tie(nullptr)->sync_with_stdio(false);
    
    auto solve = [&]() {
        int n, m;
        cin >> n >> m;

        vector<string> a(n);
        for (int i = 0; i < n; i++) {
            cin >> a[i];
        }

        vector<pair<int, int>> q{{0, 0}};
        vector vis(n, vector<int>(m));
        vector prv(n, vector<pair<int, int>>(m, {-1, -1}));
        vis[0][0] = true;

        for (int i = 0; i < int(q.size()); i++) {
            auto [x, y] = q[i];
            for (auto [dx, dy] : {pair{1, 0}, {0, 1}}) {
                int nx = x + dx, ny = y + dy;
                if (0 <= nx && nx < n && 0 <= ny && ny < m && !vis[nx][ny] && a[nx][ny] != '#') {
                    vis[nx][ny] = true;
                    prv[nx][ny] = {x, y};
                    q.emplace_back(nx, ny);
                }
            }
        }

        if (!vis[n - 1][m - 1]) {
            cout << "NIE\n";
            return;
        }

        cout << "TAK\n";

        vector<pair<int, int>> path;
        for (int x = n - 1, y = m - 1; pair(x, y) != pair(-1, -1); tie(x, y) = prv[x][y]) {
            assert(a[x][y] != '#');
            path.emplace_back(x, y);
        }

        reverse(path.begin(), path.end());
        
        string r(n, 'N'), c(m, 'N'), mv;

        for (int i = 0; i + 1 < int(path.size()); i++) {
            auto [x, y] = path[i];
            auto [nx, ny] = path[i + 1];
            if (nx - x == 1) {
                assert(y == ny);
                if (a[nx][ny] != '#') {
                    if (c[y] == 'T') {
                       a[nx][ny] ^= 'O' ^ '@';
                    }
                } 
                if (a[nx][ny] == '@') {
                    r[nx] = 'T';
                }
                mv += 'D';
            } else if (ny - y == 1) {
                assert(x == nx);
                if (a[nx][ny] != '#') {
                    if (r[x] == 'T') {
                        a[nx][ny] ^= 'O' ^ '@';
                    }
                }
                if (a[nx][ny] == '@') {
                    c[ny] = 'T';
                }
                mv += 'P';
            } else {
                assert(false);
            }
        }

        if (a[0][0] == '@') {
            if (r[0] == 'N') {
                r[0] = 'T';
            } else if (c[0] == 'N') {
                c[0] = 'T';
            } else {
                assert(false);
            }
        }

        cout << r << '\n' << c << '\n' << mv << '\n';
    };

    int t;
    cin >> t;

    while (t--) {
        solve();
    }

    return 0;
}


Details

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Test #1:

score: 0
Wrong Answer
time: 291ms
memory: 103720kb

input:

486
4 5
..#..
@@O@@
##@#O
..@.@
2 2
OO
OO
2 2
@@
@@
2 2
@@
#@
2 2
@O
#@
2 2
@@
OO
2 2
O#
@O
2 2
@#
#@
2 2
@.
.@
2 2
@#
.O
2 2
OO
.O
10 10
@O@O#O@@@#
OO#@#@@#OO
#@#@#O##O@
OO##@@O#@O
O##@@#@O#@
OO@OO@@@O@
@O#@#@O#@O
@OOOOO@##.
O@OOO##O@@
OO@@OOOO#@
10 10
@@#OOO#O@@
#@@OO@@.O@
#.O@@O#@@O
OO@@#O@#O@
.#...

output:

TAK
NTNN
NNTNT
DPPDDPP
TAK
NN
NN
DP
TAK
TT
NN
DP
TAK
TN
NT
PD
TAK
TT
NN
PD
TAK
TN
NN
DP
TAK
NT
NT
DP
NIE
TAK
TN
NT
DP
TAK
TN
NN
DP
TAK
NN
NN
DP
NIE
TAK
TNNTNNNNTN
NTNNNTNNTT
PDDDPPDDDDPPDDPPPP
TAK
NTTTNTNNNN
NTTTNTNTTN
DDDDDPDDDPPPPPDPPP
TAK
NNNTNTTTNT
NNTTTTTNNT
DDDPPDPDDDDDPPPPPP
TAK
NNNTNNNTNT
NN...

result:

wrong answer you dead (test case 4)