分析:
矩阵树定理,参见https://blog.csdn.net/qq_34454069/article/details/78035152
但这道题要用到高精度,如果一昧地使用矩阵树,可能会被卡精度,而且代码极其复杂。
所以观察一下性质,设
为n个点的生成树总数,满足
其证明就是手算矩阵的过程:
首先根据矩阵树定理,删去任意一行任意一列(即第一行与第一列)
现在根据行列式的性质,我们交换1、2行,2、3行,3、4行……
于是变成了这个样子:
现在我们发现,除了最后两行,其余的只剩下上弦三角形了。
所以要消去这两行。
尝试去模拟一下消的过程,你就会得到这个递推式了。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<iomanip>
#include<vector>
#define SF scanf
#define PF printf
#define MAXN 110
using namespace std;
int n;
typedef long long ll;
struct big {
static const int BASE = 10000; // µ±Ö»Ê¹ÓüӼõ·¨Ê±¿ÉÒÔ½ÚÊ¡¿Õ¼ä£¬°Ñ
static const int WIDTH = 4; // BASE ¸Ä³É 100000000£¬WIDTH ¸Ä³É 8
std::vector<int> s;
big(long long num = 0) { *this = num; }
big(const std::string& str) { *this = str; }
big operator = (long long num) {
s.clear();
do {
s.push_back(num % BASE);
num /= BASE;
} while (num > 0);
return *this;
}
big operator = (const std::string& str) {
s.clear();
int x, len = (str.length() - 1) / WIDTH + 1;
for (int i = 0; i < len; i++) {
int end = str.length() - i * WIDTH;
int start = std::max(0, end - WIDTH);
sscanf(str.substr(start, end - start).c_str(), "%d", &x);
s.push_back(x);
}
return *this;
}
bool operator < (const big& b) const {
if (s.size() != b.s.size()) return s.size() < b.s.size();
for (int i = s.size() - 1; i >= 0; i--)
if (s[i] != b.s[i]) return s[i] < b.s[i];
return false;
}
bool operator > (const big& b) const { return b < *this; }
bool operator <= (const big& b) const { return !(b < *this); }
bool operator >= (const big& b) const { return !(*this < b); }
bool operator != (const big& b) const { return b < *this || *this < b; }
bool operator == (const big& b) const { return !(b < *this) && !(*this < b); }
big operator + (const big& b) const {
big c;
c.s.clear();
for (int i = 0, g = 0; ; i++) {
if (g == 0 && i >= s.size() && i >= b.s.size()) break;
int x = g;
if (i < s.size()) x += s[i];
if (i < b.s.size()) x += b.s[i];
c.s.push_back(x % BASE);
g = x / BASE;
}
return c;
}
big operator - (const big& b) const {
big c;
c.s.clear();
if (*this == b) {
c.s.push_back(0);
} else if (*this < b) {
for (int i = 0, g = 0; ; i++) {
if (g == 0 && i >= b.s.size() && i >= s.size()) {
c.s[i - 1] = -c.s[i - 1]; break;
}
int x = g;
if (i < b.s.size()) x += b.s[i];
if (i < s.size()) x -= s[i];
if (x < 0) {
c.s.push_back(BASE + x % BASE);
g = x / BASE - 1;
} else {
c.s.push_back(x % BASE);
g = x / BASE;
}
}
} else {
for (int i = 0, g = 0; ; i++) {
if (g == 0 && i >= s.size() && i >= b.s.size()) break;
int x = g;
if (i < s.size()) x += s[i];
if (i < b.s.size()) x -= b.s[i];
if (x < 0) {
c.s.push_back(BASE + x % BASE);
g = x / BASE - 1;
} else {
c.s.push_back(x % BASE);
g = x / BASE;
}
}
}
return c;
}
big operator * (const big& b) const {
big c; int len = s.size() + b.s.size(); bool flag = false;
c.s.clear();
if (*this == 0 || b == 0) {c.s.push_back(0); return 0;}
if (*this < 0 && b > 0) flag = true;
if (*this > 0 && b < 0) flag = true;
for (int i = 0, g = 0; ; i++) {
if (g == 0 && i >= len) break;
int x = g;
for (int u = 0, v = i, temp; v >= 0; u++, v--)
if (u < s.size() && v < b.s.size()) {
temp = s[u] * b.s[v];
if (temp < 0) temp = -temp;
x += temp;
}
c.s.push_back(x % BASE);
g = x / BASE;
}
for (int i = c.s.size() - 1; i >= 0 && c.s[i] == 0; i--)
c.s.pop_back();
if (flag) c.s[c.s.size() - 1] = -c.s[c.s.size() - 1];
return c;
}
inline void killzero() {
while (s.back() == 0 && s.size() > 1) s.pop_back();
}
inline void reverse() {
int len = s.size();
for (int i = 0; i < len >> 1; ++i) swap(s[i], s[len - i - 1]);
}
big operator / (const big &b) const {
big c, t;
c.s.clear(); t.s.clear();
for (int i = s.size() - 1; i >= 0; --i) {
t.s.push_back(s[i]);
int x = 0;
while (b <= t) { t -= b; x++; }
c.s.push_back(x);
}
c.reverse();
c.killzero();
return c;
}
big operator += (const big& b) {
*this = *this + b; return *this;
}
big operator ++ (int) {
*this = *this + 1; return *this;
}
big operator ++ () {
*this = *this + 1; return *this;
}
big operator -= (const big& b) {
*this = *this - b; return *this;
}
big operator -- (int) {
*this = *this - 1; return *this;
}
big operator -- () {
*this = *this - 1; return *this;
}
big operator *= (const big& b) {
*this = *this * b; return *this;
}
big operator /= (const big& b) {
*this = *this / b; return *this;
}
// big operator %= (const big& b) {
// *this = *this % b; return *this;
// }
};
std::ostream& operator << (std::ostream& out, const big& x) {
out << x.s.back();
for (int i = x.s.size() - 2; i >= 0; i--) {
char buf[20];
sprintf(buf, "%04d", x.s[i]);
for (int j = 0; j < strlen(buf); j++) out << buf[j];
}
return out;
}
big a[MAXN];
int main(){
SF("%d",&n);
a[1]=1;
a[0]=0;
for(int i=2;i<=n;i++){
a[i]=a[i-1]*3-a[i-2]+2;
}
cout<<a[n];
}