CF917D. Stranger Trees & TopCoder13369. TreeDistance（变元矩阵树定理＋高斯消元）
题目链接
CF917D：https://codeforces.com/problemset/problem/917/D
TopCoder13369：https://community.topcoder.com/stat?c=problem_statement&pm=13369
题解
首先分析 CF917D。
我们考虑能否将树上的边的贡献特殊表现出来。
记原树为 \(T\)，我们构造一幅 \(n\) 个结点的无向完全图，并设置一个值 \(x\)，对于无向边 \((u, v)\)，其权值 \(w_{(u, v)}\) 满足：
\[w_{(u, v)} = \begin{cases} x, & (u, v) \in T \\ 1, & (u, v) \notin T \end{cases}
\]
如果我们令无向完全图的所有生成树中，恰好有 \(i\) 条边属于原树 \(T\) 的生成树数量为 \({\rm ans}_i\)，那么不难发现，根据变元矩阵树定理（基尔霍夫矩阵的任意 \(n - 1\) 阶主子式的行列式的绝对值即为所有生成树的边权积之和），求得的基尔霍夫矩阵的 \(n - 1\) 阶主子式的值即为 \(\sum_\limits{i = 0}^{n - 1} x^i {\rm ans}_i\)，因为若一个生成树包含了 \(i\) 条原树边，那么该生成树的边权积即为 \(x^i\)。
如果我们将 \({\rm ans}_i\) 视为未知数，那么我们就可以通过枚举 \(x: 0 \sim n - 1\) 来得到一个 \(n\) 元线性方程组。直接高斯消元即可求得所有的 \({\rm ans}_i\)。
接下来分析 TopCoder13369。
考虑原树 \(T\) 最少能通过多少次操作得到另一棵树 \(T'\)，显然若在原树 \(T\) 中出现而在树 \(T'\) 中没有出现的边数为 \(w\)，那么只需要 \(w\) 次操作即可将原树 \(T\) 变为树 \(T'\)。因此答案即为有 \(i(i \geq n - 1 - k)\) 条边属于原树的生成树数量，即 \(\sum_\limits{i = n - 1 - k}^{n - 1} {\rm ans}_i\)。其余部分与上题相同。
两题的时间复杂度均为 \(O(n^4)\)。
代码
CF917D 代码如下：
#include<bits/stdc++.h>
using namespace std;
const int N = 105, mod = 1e9 + 7;
void add(int& x, int y) {
x += y;
if (x >= mod) {
x -= mod;
}
}
void sub(int& x, int y) {
x -= y;
if (x < 0) {
x += mod;
}
}
int mul(int x, int y) {
return (long long) x * y % mod;
}
int qpow(int v, int p) {
int result = 1;
for (; p; p >>= 1, v = mul(v, v)) {
if (p & 1) {
result = mul(result, v);
}
}
return result;
}
int n, a[N][N], b[N][N];
bool old[N][N];
int matrix_tree() {
int result = 1;
for (int i = 2; i <= n; ++i) {
int rev = i;
for (int j = i + 1; j <= n; ++j) {
if (a[j][i]) {
rev = j;
break;
}
}
if (rev != i) {
result = (mod - result) % mod;
for (int j = i; j <= n; ++j) {
swap(a[i][j], a[rev][j]);
}
}
for (int j = i + 1; j <= n; ++j) {
int p = mul(a[j][i], qpow(a[i][i], mod - 2));
for (int k = i; k <= n; ++k) {
sub(a[j][k], mul(p, a[i][k]));
}
}
result = mul(result, a[i][i]);
}
return result;
}
int main() {
scanf("%d", &n);
for (int i = 1; i < n; ++i) {
int u, v;
scanf("%d%d", &u, &v);
old[u][v] = old[v][u] = true;
}
for (int x = 0; x < n; ++x) {
memset(a, 0, sizeof a);
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= n; ++j) {
int v = old[i][j] ? x : 1;
add(a[j][j], v);
sub(a[i][j], v);
}
}
for (int i = 0, v = 1; i < n; ++i, v = mul(v, x)) {
b[x][i] = v;
}
b[x][n] = matrix_tree();
}
for (int i = 0; i < n; ++i) {
int rev = i;
for (int j = i + 1; j < n; ++j) {
if (b[j][i]) {
rev = j;
break;
}
}
if (rev != i) {
for (int j = i; j <= n; ++j) {
swap(b[i][j], b[rev][j]);
}
}
for (int j = i + 1; j < n; ++j) {
int p = mul(b[j][i], qpow(b[i][i], mod - 2));
for (int k = i; k <= n; ++k) {
sub(b[j][k], mul(p, b[i][k]));
}
}
}
for (int i = n - 1; ~i; --i) {
for (int j = i + 1; j < n; ++j) {
sub(b[i][n], mul(b[i][j], b[j][n]));
}
b[i][n] = mul(b[i][n], qpow(b[i][i], mod - 2));
}
for (int i = 0; i < n; ++i) {
printf("%d%c", b[i][n], " \n"[i == n - 1]);
}
return 0;
}
TopCoder13369 代码如下：
直接粘过来改一下即可。
#include<bits/stdc++.h>
using namespace std;
const int N = 55, mod = 1e9 + 7;
void add(int& x, int y) {
x += y;
if (x >= mod) {
x -= mod;
}
}
void sub(int& x, int y) {
x -= y;
if (x < 0) {
x += mod;
}
}
int mul(int x, int y) {
return (long long) x * y % mod;
}
int qpow(int v, int p) {
int result = 1;
for (; p; p >>= 1, v = mul(v, v)) {
if (p & 1) {
result = mul(result, v);
}
}
return result;
}
int n, a[N][N], b[N][N];
bool old[N][N];
int matrix_tree() {
int result = 1;
for (int i = 1; i < n; ++i) {
int rev = i;
for (int j = i + 1; j < n; ++j) {
if (a[j][i]) {
rev = j;
break;
}
}
if (rev != i) {
result = (mod - result) % mod;
for (int j = i; j < n; ++j) {
swap(a[i][j], a[rev][j]);
}
}
for (int j = i + 1; j < n; ++j) {
int p = mul(a[j][i], qpow(a[i][i], mod - 2));
for (int k = i; k < n; ++k) {
sub(a[j][k], mul(p, a[i][k]));
}
}
result = mul(result, a[i][i]);
}
return result;
}
class TreeDistance {
public:
int countTrees(vector<int> p, int c) {
n = p.size() + 1;
for (int i = 0; i < p.size(); ++i) {
old[p[i]][i + 1] = old[i + 1][p[i]] = true;
}
for (int x = 0; x < n; ++x) {
memset(a, 0, sizeof a);
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
int v = old[i][j] ? x : 1;
add(a[j][j], v);
sub(a[i][j], v);
}
}
for (int i = 0, v = 1; i < n; ++i, v = mul(v, x)) {
b[x][i] = v;
}
b[x][n] = matrix_tree();
}
for (int i = 0; i < n; ++i) {
int rev = i;
for (int j = i + 1; j < n; ++j) {
if (b[j][i]) {
rev = j;
break;
}
}
if (rev != i) {
for (int j = i; j <= n; ++j) {
swap(b[i][j], b[rev][j]);
}
}
for (int j = i + 1; j < n; ++j) {
int p = mul(b[j][i], qpow(b[i][i], mod - 2));
for (int k = i; k <= n; ++k) {
sub(b[j][k], mul(p, b[i][k]));
}
}
}
for (int i = n - 1; ~i; --i) {
for (int j = i + 1; j < n; ++j) {
sub(b[i][n], mul(b[i][j], b[j][n]));
}
b[i][n] = mul(b[i][n], qpow(b[i][i], mod - 2));
}
int answer = 0;
for (int i = n - 1; i >= max(0, n - 1 - c); --i) {
add(answer, b[i][n]);
}
return answer;
}
};
/*
TreeDistance solver;
int main() {
int n, k;
scanf("%d", &n);
vector<int> p(n - 1);
for (int i = 0; i < n - 1; ++i) {
scanf("%d", &p[i]);
}
scanf("%d", &k);
printf("%d\n", solver.countTrees(p, k));
return 0;
}
*/
