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#185282#5407. 基础图论练习题LCX7560 22ms8284kbC++234.2kb2023-09-21 20:37:472023-09-21 20:37:48

Judging History

你现在查看的是最新测评结果

  • [2023-09-21 20:37:48]
  • 评测
  • 测评结果:0
  • 用时:22ms
  • 内存:8284kb
  • [2023-09-21 20:37:47]
  • 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];
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 u : vec)
        for (int v : vec) if (e[u][v]) deg[u]++;
    function <void(int)> dfs = [&] (int u) {
        if (vis[u]) return;
        vis[u] = 1, a.push_back(u);
        for (int v : vec) if (e[u][v]) dfs(v);
    }; dfs(vec[0]);
    int m = a.size();
    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 - 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;
        }
    }
}

void work() {
    scanf("%d", &n);
    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;
            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);
    //         ans[u][v] = ans[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;
}

详细

Subtask #1:

score: 0
Wrong Answer

Test #1:

score: 0
Wrong Answer
time: 22ms
memory: 8284kb

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:

78571499
13
251125658
298648245
50
0
1983
2
234841943
224784877
764560207
29948
2
225520384
14
2065151
2502
0
63721755
512437818
2080617
35834
30719
2556784
894125799
360162342
316822991
79011162
149138696
63494
263454681
822310163
32717
2088694
578624420
472665248
18
45
95
1046466
230645497
13
2524...

result:

wrong answer 1st numbers differ - expected: '281603732', found: '78571499'

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%