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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
#447574#7677. Easy Diameter ProblemzhaohaikunWA 2ms5236kbC++204.5kb2024-06-18 16:40:342024-06-18 16:40:35

Judging History

你现在查看的是最新测评结果

  • [2024-06-18 16:40:35]
  • 评测
  • 测评结果:WA
  • 用时:2ms
  • 内存:5236kb
  • [2024-06-18 16:40:34]
  • 提交

answer

// MagicDark
#include <bits/stdc++.h>
#define debug cerr << "\033[32m[" << __LINE__ << "]\033[0m "
#define SZ(x) ((int) x.size() - 1)
#define all(x) x.begin(), x.end()
#define ms(x, y) memset(x, y, sizeof x)
#define F(i, x, y) for (int i = (x); i <= (y); i++)
#define DF(i, x, y) for (int i = (x); i >= (y); i--)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
template <typename T> T& chkmax(T& x, T y) {return x = max(x, y);}
template <typename T> T& chkmin(T& x, T y) {return x = min(x, y);}
template <typename T> T& read(T &x) {
	x = 0; int f = 1; char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = - f;
	for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
	return x *= f;
}
const int N = 305, MOD = 1e9 + 7;
inline int& add(int &x, ll y) {return x = (x + y) % MOD;}
int n, x[N], y[N], cnt, d[N][N], mx, c, fa[N], e[N][N], f[2][N][N], ans, C[N][N], g[N][N];
vector <int> v[N];
int fac[N], ifac[N], inv[N];
inline void binom(int x) {
	fac[0] = ifac[0] = inv[1] = 1;
	F(i, 2, x) inv[i] = (ll) (MOD - MOD / i) * inv[MOD % i] % MOD;
	F(i, 1, x) fac[i] = (ll) fac[i - 1] * i % MOD, ifac[i] = (ll) ifac[i - 1] * inv[i] % MOD;
	F(i, 0, x) {
		C[i][0] = 1;
		F(j, 1, i) C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
	}
}
// inline int C(int x, int y) {return x < y || y < 0 ? 0 : (ll) fac[x] * ifac[y] % MOD * ifac[x - y] % MOD;}
void dfs(int id, int x, int fa, int dep) {
	// debug << x << " " << id << " " << dep << endl;
	::fa[x] = fa;
	if (dep + 1 > mx) {
		mx = dep + 1;
		int w = x;
		F(i, 1, mx >> 1) w = ::fa[w];
		if (mx & 1) c = e[w][::fa[w]];
		else c = w;
	}
	d[id][dep]++;
	for (int i: v[x])
		if (i != fa) dfs(id, i, x, dep + 1);
}
signed main() {
	binom(read(n));
	if (n == 1) {
		cout << 1;
		return 0;
	}
	F(i, 1, n - 1) {
		y[i << 1 | 1] = read(x[i << 1]), x[i << 1 | 1] = read(y[i << 1]);
		e[x[i << 1]][y[i << 1]] = i << 1;
		e[y[i << 1]][x[i << 1]] = i << 1 ^ 1;
		v[x[i << 1]].push_back(y[i << 1]);
		v[y[i << 1]].push_back(x[i << 1]);
	}
	F(i, 2, 2 * n - 1) dfs(i, y[i], x[i], 0);
	// debug << mx << " " << c << endl;
	if (mx & 1) {
		int a = d[c][mx >> 1], b = d[c ^ 1][mx >> 1];
		F(i, 0, a - 1) f[mx & 1 ^ 1][c][i] = (ll) fac[b] * C[i + b - 1][i] % MOD;
		F(i, 0, b - 1) f[mx & 1 ^ 1][c ^ 1][i] = (ll) fac[a] * C[i + a - 1][i] % MOD;
	} else {
		for (int i: v[c]) {
			int a = d[e[c][i]][(mx >> 1) - 1], b = d[e[c][i] ^ 1][mx >> 1];
			// debug << a << " " << b << endl;
			// debug << e[c][i] << endl;
			F(j, 0, a - 1) f[mx & 1 ^ 1][e[c][i]][j] = (ll) fac[b] * C[j + b - 1][j] % MOD;
		}
	}
	DF(i, mx - 1, 1) {
		int cur = i & 1, to = cur ^ 1;
		if (i & 1) {
			F(j, 2, n * 2 - 1) {
				int a = d[j][i >> 1], b = d[j ^ 1][i >> 1];
				if (!a || !b) continue;
				F(k, 0, a - 1) {
					g[k][0] = ((k ? g[k - 1][0] : 0) + f[cur][j][k]) % MOD;
					// if (f[cur][j][k]) debug << i << ' ' << j << " " << k << " " << f[cur][j][k] << endl;
					f[cur][j][k] = 0;
					F(l, 1, b - 1) g[k][l] = (g[k][l - 1] + (k ? g[k - 1][l] : 0)) % MOD;
					add(f[to][j][k], (ll) g[k][b - 1] * fac[b]);
					// F(l, 0, b - 1) add(f[to][j ^ 1][l], (ll) f[cur][j][k] * C[a - k - 1 + l][l] % MOD * fac[b]);
				}
				F(k, 0, b - 1) add(f[to][j ^ 1][k], (ll) g[a - 1][k] * fac[a]);
			}
		} else {
			F(j, 2, n * 2 - 1) {
				int a = d[j][i >> 1], b = d[j ^ 1][(i >> 1) - 1];
				if (!a || !b) continue;
				// debug << i << " " << j << " " << a << " " << b << endl;
				F(k, 0, a - 1) {
					// if (f[cur][j][k] != 1 || k) f[cur][j][k] = 0;
					g[k][0] = ((k ? g[k - 1][0] : 0) + f[cur][j][k]) % MOD;
					// if (f[cur][j][k]) debug << i << ' ' << j << " " << k << " " << f[cur][j][k] << endl;
					f[cur][j][k] = 0;
					F(l, 1, b) g[k][l] = (g[k][l - 1] + (k ? g[k - 1][l] : 0)) % MOD;
				}
				// debug << i << " " << j << endl;
				F(k, 0, b - 1) add(f[to][j ^ 1][k], (ll) g[a - 1][k] * fac[a]);//, debug << "! " << k << " " << g[a - 1][k] << " " << fac[a] << endl;
				// continue;
				// debug << y[j] << endl;
				for (int k: v[y[j]]) {
					int c = d[e[y[j]][k]][(i >> 1) - 1];
					if (k != x[j] && c) {
						F(l, 0, c - 1) {
							add(f[to][e[y[j]][k]][l], (ll) g[a - c + l][b - 1] * C[a - c + l][l] % MOD * fac[b] % MOD * fac[a - c]);
							if (a - c + l) add(f[to][e[y[j]][k]][l], (ll) g[a - c + l - 1][b] * C[a - c + l - 1][l] % MOD * fac[b] % MOD * fac[a - c]);
						}
					}
				}
			}
		}
	}
	F(i, 2, n * 2 - 1) add(ans, f[0][i][0]);
	cout << ans;
	return 0;
}
/* why?
*/

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 3632kb

input:

3
1 2
3 2

output:

4

result:

ok 1 number(s): "4"

Test #2:

score: 0
Accepted
time: 0ms
memory: 3704kb

input:

5
4 1
4 5
1 2
1 3

output:

28

result:

ok 1 number(s): "28"

Test #3:

score: 0
Accepted
time: 1ms
memory: 3696kb

input:

7
5 7
2 5
2 1
1 6
3 6
4 1

output:

116

result:

ok 1 number(s): "116"

Test #4:

score: 0
Accepted
time: 1ms
memory: 4704kb

input:

100
89 60
66 37
59 74
63 49
69 37
9 44
94 22
70 30
79 14
25 33
23 4
78 43
63 30
95 7
6 59
56 31
21 56
58 43
95 28
81 19
63 40
58 87
81 38
100 55
78 23
37 78
86 70
33 11
52 17
42 19
19 13
70 100
97 79
66 67
66 27
82 55
15 27
68 33
97 26
46 20
70 93
1 10
69 79
81 45
58 95
24 63
79 51
85 11
12 43
41 64...

output:

748786623

result:

ok 1 number(s): "748786623"

Test #5:

score: -100
Wrong Answer
time: 2ms
memory: 5236kb

input:

300
109 123
221 101
229 22
114 101
110 258
50 79
26 1
238 47
140 271
77 213
140 125
19 179
111 96
194 114
57 101
1 251
19 13
92 238
114 116
193 140
153 238
1 173
252 220
1 22
124 197
196 214
196 224
1 138
201 90
184 56
266 154
71 184
46 15
256 27
1 69
1 32
22 182
237 196
111 1
279 91
139 196
114 80
...

output:

0

result:

wrong answer 1st numbers differ - expected: '47013557', found: '0'