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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
#313488#6331. Jewel GamePlentyOfPenaltyWA 4ms17528kbC++203.9kb2024-01-24 19:53:582024-01-24 19:53:58

Judging History

你现在查看的是最新测评结果

  • [2024-01-24 19:53:58]
  • 评测
  • 测评结果:WA
  • 用时:4ms
  • 内存:17528kb
  • [2024-01-24 19:53:58]
  • 提交

answer

#include <bits/stdc++.h>
#define STATE [(1 << K) + 10][N + 10][N + 10]
#define all(x) begin(x), end(x)
using namespace std;
const int N = 30, K = 10;
const int INF = 1e9;
struct state {
    int m, p, q, val;
} t;
int n, m, a, b, k, u, v, pos[K + 10], w[K + 10], id[N + 10];
int dp STATE, vis STATE, d STATE, mx;
vector<int> e[N + 10];
vector<state> ut STATE;
queue<state> q;
int main() {
#ifdef popteam
    freopen("O.in", "r", stdin);
    freopen("O.out", "w", stdout);
#endif
    cin.tie(0)->sync_with_stdio(0);
    cin >> n >> m >> a >> b;
    for (int i = 1; i <= m; ++i)
        cin >> u >> v, e[u].push_back(v);
    cin >> k;
    for (int i = 1; i <= k; ++i) {
        cin >> pos[i] >> w[i];
        id[pos[i]] = i;
    }
    for (int i = 1; i < (1 << k); ++i) {
        for (int j = 1; j <= n; ++j)
            for (int k = 1; k <= n; ++k) {
                for (int t : e[j]) {
                    if (id[t] && ((i >> id[t] - 1) & 1)) {
                        ut[i - (1 << id[t] - 1)][k][t].push_back((state){i, j, k, w[id[t]]});
                        // cout << "(" << i << " " << j << " " << k << ")->(" << i - (1 << id[t] - 1) << " " << k << " " << t << ")\n";
                        ++d[i][j][k];
                    } else {
                        ut[i][k][t].push_back((state){i, j, k, 0});
                        // cout << "(" << i << " " << j << " " << k << ")->(" << i << " " << k << " " << t << ")\n";
                        ++d[i][j][k];
                    }
                }
            }
    }
    for (int i = 1; i < (1 << k); ++i)
        for (int j = 1; j <= n; ++j)
            for (int k = 1; k <= n; ++k)
                dp[i][j][k] = -INF;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j) {
            vis[0][i][j] = 1, dp[0][i][j] = 0;
            for (state x : ut[0][i][j]) {
                --d[x.m][x.p][x.q], dp[x.m][x.p][x.q] = max(dp[x.m][x.p][x.q], x.val - dp[0][i][j]);
            }
        }
    for (int i = 1; i < (1 << k); ++i) {
        for (int j = 1; j <= n; ++j)
            for (int k = 1; k <= n; ++k)
                if (!d[i][j][k]) q.push((state){i, j, k});
        while (1) {
            while (!q.empty()) {
                t = q.front(), q.pop();
                vis[t.m][t.p][t.q] = 1;
                // cout << "DP " << t.m << " " << t.p << " " << t.q << "=" << dp[t.m][t.p][t.q] << "\n";
                for (state x : ut[t.m][t.p][t.q]) {
                    --d[x.m][x.p][x.q], dp[x.m][x.p][x.q] = max(dp[x.m][x.p][x.q], x.val - dp[t.m][t.p][t.q]);
                    if (x.m == i && !d[x.m][x.p][x.q]) q.push(x);
                }
            }
            mx = 0;
            for (int j = 1; j <= n; ++j)
                for (int k = 1; k <= n; ++k)
                    if (!vis[i][j][k] && dp[i][j][k] > mx) {
                        mx = dp[i][j][k], t = (state){i, j, k};
                    }
            if (mx > 0) {
                vis[t.m][t.p][t.q] = 1;
                // cout << "DP " << t.m << " " << t.p << " " << t.q << "=" << dp[t.m][t.p][t.q] << "\n";
                for (state x : ut[t.m][t.p][t.q]) {
                    --d[x.m][x.p][x.q], dp[x.m][x.p][x.q] = max(dp[x.m][x.p][x.q], x.val - dp[t.m][t.p][t.q]);
                    if (x.m == i && !d[x.m][x.p][x.q]) q.push(x);
                }
            } else
                break;
        }
        for (int j = 1; j <= n; ++j)
            for (int k = 1; k <= n; ++k)
                if (!vis[i][j][k]) {
                    vis[i][j][k] = 1, dp[i][j][k] = 0;
                    // cout << "DP " << i << " " << j << " " << k << "=0\n";
                    for (state x : ut[i][j][k]) {
                        --d[x.m][x.p][x.q], dp[x.m][x.p][x.q] = max(dp[x.m][x.p][x.q], x.val - dp[i][j][k]);
                    }
                }
    }
    cout << dp[(1 << k) - 1][a][b] << "\n";
}

詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 7936kb

input:

5 16 1 1
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 2
3 4
3 5
4 2
4 3
4 5
5 2
5 3
5 4
4
2 4
3 84
4 38
5 96

output:

46

result:

ok 1 number(s): "46"

Test #2:

score: 0
Accepted
time: 0ms
memory: 8892kb

input:

8 16 8 4
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 1
1 5
2 6
3 7
4 8
5 1
6 2
7 3
8 4
6
1 29
2 34
3 41
5 7
6 26
7 94

output:

-23

result:

ok 1 number(s): "-23"

Test #3:

score: 0
Accepted
time: 3ms
memory: 7768kb

input:

5 5 2 1
1 1
2 3
3 4
4 5
5 2
2
4 1000000
5 100000

output:

1100000

result:

ok 1 number(s): "1100000"

Test #4:

score: -100
Wrong Answer
time: 4ms
memory: 17528kb

input:

10 20 1 2
1 4
1 7
2 2
2 4
3 6
3 3
4 8
4 7
5 7
5 1
6 9
6 2
7 9
7 3
8 8
8 6
9 7
9 8
10 10
10 2
8
3 92067840
4 2874502
5 36253165
6 70758738
7 4768969
8 16029185
9 16207515
10 44912151

output:

164899375

result:

wrong answer 1st numbers differ - expected: '132484345', found: '164899375'