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#185290#5407. 基础图论练习题LCX7560 0ms0kbC++144.3kb2023-09-21 20:43:162023-09-21 20:43:17

Judging History

你现在查看的是最新测评结果

  • [2023-09-21 20:43:17]
  • 评测
  • 测评结果:0
  • 用时:0ms
  • 内存:0kb
  • [2023-09-21 20:43:16]
  • 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 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;
                if (ans[u][v] != ans2[u][v]) assert(0);
            }
        }
    }
}

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);
            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
Runtime Error

Test #1:

score: 0
Runtime Error

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:


result:


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%