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#185314#5407. 基础图论练习题LCX7560 25ms7884kbC++144.6kb2023-09-21 21:12:092023-09-21 21:12:10

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你现在查看的是最新测评结果

  • [2023-09-21 21:12:10]
  • 评测
  • 测评结果:0
  • 用时:25ms
  • 内存:7884kb
  • [2023-09-21 21:12:09]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef double db;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
#define fir first
#define sec second
typedef vector <int> vi;
typedef vector <ll> vl;

#ifdef LCX
#define msg(args...) fprintf(stderr, args)
#else
#define msg(...) void()
#endif

constexpr int maxn = 5010, mod = 1e9 + 7;
int n, ans[maxn][maxn], ans2[maxn][maxn];
bitset <maxn> e[maxn];

int hextodec(char ch) {
    if (isalpha(ch)) return ch - 'A' + 10;
    return ch - '0';
}

int dfn[maxn], low[maxn], idx, stk[maxn], tp, ins[maxn], scc[maxn], cnt_scc, siz[maxn];
void dfs(int u) {
    dfn[u] = low[u] = ++idx;
    stk[++tp] = u, ins[u] = 1;
    for (int v = 1; v <= n; ++v) if (e[u][v]) {
        if (!dfn[v]) dfs(v), low[u] = min(low[u], low[v]);
        else if (ins[v]) low[u] = min(low[u], dfn[v]);
    }
    if (dfn[u] == low[u]) {
        int x;
        ++cnt_scc;
        do {
            x = stk[tp--];
            ins[x] = 0;
            scc[x] = cnt_scc;
            siz[cnt_scc]++;
        } while (x != u);
    }
}
int vis[maxn], deg[maxn], sum[maxn], p[maxn];
void solve(int c) {
    vi vec, a;
    for (int i = 1; i <= n; ++i)
        if (scc[i] == c) vec.push_back(i), vis[i] = 0, deg[i] = 0;
    // for (int x : vec) msg("%d ", x);
    // msg("\n");
    for (int u : vec)
        for (int v : vec) if (e[u][v]) deg[u]++;
    int m = vec.size();
    for (int x : vec) {
        a.push_back(x);
        int i = a.size() - 1;
        while (1) {
            int flg = 1;
            if (i) flg &= e[a[i - 1]][a[i]];
            if (i < (int) a.size() - 1) flg &= e[a[i]][a[i + 1]];
            if (flg) break;
            swap(a[i], a[i - 1]), --i;
        }
    }
    for (int i = 0; i < m; ++i) vis[a[i]] = i;
    auto calc = [&] (int x, int y) {
        deg[x]--, deg[y]++;
        for (int i = 0; i <= m; ++i) sum[i] = 0;
        for (int x : vec) sum[deg[x]]++;
        for (int i = 1; i <= m; ++i) sum[i] += sum[i - 1];
        for (int x : vec) p[sum[x]--] = deg[x];
        int s = 0, cnt = 0;
        for (int i = 1; i <= m; ++i)
            s += p[i], cnt += (s == i * (i - 1) / 2);
        ans[x][y] = ans[y][x] = cnt - 1 + cnt_scc;
        deg[x]++, deg[y]--;
    };
    int l = m - 1;
    // for (int i = 0; i < m; ++i) msg("%d%c", a[i], " \n"[i == m - 1]);
    for (int i = 0; i < m - 1; ++i) calc(a[i], a[i + 1]);
    for (int i = m - 1; i >= 0; --i) {
        int u = a[i];
        for (int v : vec) if (e[u][v]) {
            if (l > vis[v]) calc(u, v), l = vis[v];
            else if (vis[v] != i + 1) {
                ans[u][v] = ans[v][u] = cnt_scc;
                if (ans[u][v] != ans2[u][v]) assert(0);
            }
        }
    }
}

mt19937 rnd(time(0));

void work() {
    // scanf("%d", &n);
    n = 5;
    for (int i = 2; i <= n; ++i) {
        // static char s[maxn];
        // scanf("%s", s);
        for (int j = 1; j < i; ++j) {
            // int c = (hextodec(s[(j - 1) / 4]) >> ((j - 1) % 4)) & 1;
            int c = rnd() & 1;
            e[i][j] = c, e[j][i] = !c;
        }
    }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j)
            msg("%d", (int) e[i][j]);
        msg("\n");
    }
    for (int u = 1; u <= n; ++u)
        for (int v = 1; v < u; ++v) {
            e[u].flip(v), e[v].flip(u);
            idx = tp = cnt_scc = 0;
            for (int i = 1; i <= n; ++i) dfn[i] = low[i] = ins[i] = siz[i] = 0;
            for (int i = 1; i <= n; ++i) if (!dfn[i]) dfs(i);
            e[u].flip(v), e[v].flip(u);
            ans2[u][v] = ans2[v][u] = cnt_scc;
        }
    idx = tp = cnt_scc = 0;
    for (int i = 1; i <= n; ++i) dfn[i] = low[i] = ins[i] = siz[i] = 0;
    for (int i = 1; i <= n; ++i) if (!dfn[i]) dfs(i);
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j < i; ++j)
            if (scc[i] != scc[j]) {
                if (abs(scc[i] - scc[j]) == 1 && siz[scc[i]] == 1 && siz[scc[j]] == 1)
                    ans[i][j] = ans[j][i] = cnt_scc;
                else
                    ans[i][j] = ans[j][i] = cnt_scc - abs(scc[i] - scc[j]);
            }
    for (int i = 1; i <= cnt_scc; ++i) solve(i);
    // for (int i = 1; i <= n; ++i)
    //     for (int j = 1; j <= n; ++j)
    //         printf("%d%c", ans[i][j], " \n"[j == n]);
    int pw2 = 1, res = 0;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j < i; ++j)
            res = (res + (ll) ans[i][j] * pw2) % mod, pw2 = pw2 * 2 % mod;
    printf("%d\n", res);
}
int main() {
    int T; scanf("%d", &T);
    while (T--) work();
    return 0;
}

詳細信息

Subtask #1:

score: 0
Wrong Answer

Test #1:

score: 0
Wrong Answer
time: 25ms
memory: 7884kb

input:

10000
100
1
2
2
8
C0
F0
27
78
AE1
C01
511
D87
EF20
3873
2742
73D0
DC9B0
FB2A3
9C011
9B4E0
95DC00
A7B980
F43531
6A6245
5347BE0
1A6C8A1
88E46D6
64CF3AE
D25F63C1
C894E4C3
1C0AFD73
EC1C3F9A
087CE17C0
22149A380
B28038AF1
B9CA21C7F
D78F5307C1
49045489A2
72C4DE6FD1
7713F40D05
EEE8878EEC1
310E62812B1
DA9D5B...

output:

1353
4347
2333
239
1483
2921
1267
1395
3376
833
583
498
2048
2272
1023
1827
755
281
1973
1585
1352
945
1001
2249
703
2153
453
2487
535
410
1491
454
2185
233
2048
744
830
369
1614
190
241
831
1473
1046
445
1697
2487
1014
954
1760
598
3667
174
2411
508
990
806
1617
1071
1841
2249
1794
2274
1731
1041
1...

result:

wrong answer 1st numbers differ - expected: '281603732', found: '1353'

Subtask #2:

score: 0
Skipped

Dependency #1:

0%

Subtask #3:

score: 0
Skipped

Dependency #1:

0%

Subtask #4:

score: 0
Skipped

Dependency #1:

0%

Subtask #5:

score: 0
Skipped

Dependency #1:

0%