QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#149446#4432. Jungle Trailrgnerdplayer#AC ✓251ms103564kbC++202.7kb2023-08-24 16:52:432023-08-24 16:52:46

Judging History

你现在查看的是最新测评结果

  • [2023-08-24 16:52:46]
  • 评测
  • 测评结果:AC
  • 用时:251ms
  • 内存:103564kb
  • [2023-08-24 16:52:43]
  • 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 (path[1] == pair{1, 0}) {
                r[0] = 'T';
            } else if (path[1] == pair{0, 1}) {
                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: 100
Accepted
time: 251ms
memory: 103564kb

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
NN
TT
PD
TAK
NT
TN
PD
TAK
TN
NN
DP
TAK
NT
NT
DP
NIE
TAK
TN
NT
DP
TAK
TN
NN
DP
TAK
NN
NN
DP
NIE
TAK
NNNTNNNNTN
TTNNNTNNTT
PDDDPPDDDDPPDDPPPP
TAK
NTTTNTNNNN
NTTTNTNTTN
DDDDDPDDDPPPPPDPPP
TAK
NNNTNTTTNT
NNTTTTTNNT
DDDPPDPDDDDDPPPPPP
TAK
NNNTNNNTNT
NN...

result:

ok ac (486 test cases)

Extra Test:

score: 0
Extra Test Passed