洛谷 P1411 树

最近在做些树形DP练练手

原题链接

大意就是给你一棵树，你可以断开任意数量的边，使得剩下的联通块大小乘积最大。

样例

8

1 2

1 3

2 4

2 5

3 6

3 7

6 8

输出

18

我首先想的是设\(f[i]\)表示以\(i\)为根的子树可获得的最大收益，但是会发现这样无法转移。考虑再加一维，\(f[i][j]\)表示以\(i\)的子树中，\(i\)所在的联通块大小为\(j\)的最大价值。然后我就傻了，想了半天也没想起来怎么转移，最后只好看了一眼题解。其实转移好简单的，貌似是个树上背包？考虑在\(dfs\)的过程中进行\(DP\)，每当访问完一个点\(i\)的子结点时，累加一下\(sz[i]\)，就枚举\(j\)，并且用当前子结点的\(DP\)值来更新\(f[i][j]\)。转移方程大概会长成下面这个样子：

$f[i][j]=max(f[i][j],f[i][k]*f[v][j-k])$
（理解的话，就是把之前的大小为$k$的联通块和在当前子树中大小为$j-k$的联通块拼起来）
同时，我们特别定义$f[i][0]$表示以$i$为根的子树可获得的价值，则他的转移方程比较特殊：
$f[i][0]=max(f[i][0],f[i][j]*j)$
如果到这里这道题就结束的话，代码会长成下面这样：
```cpp
#include

using namespace std;

define N 700

define ll long long

int n, eid, sz[N+5], head[N+5];

ll f[N+5][N+5];

struct Edge {

int next, to;

}e[2*N+5];

void addEdge(int u, int v) {

e[++eid].next = head[u];

e[eid].to = v;

head[u] = eid;

}

void dp(int u, int fa) {

sz[u] = 1, f[u][0] = f[u][1] = 1;

for(int i = head[u]; i; i = e[i].next) {

int v = e[i].to;

if(v == fa) continue;

dp(v, u);

sz[u] += sz[v];

for(int j = sz[u]; j >= 1; --j) { //枚举i所在的联通块大小

for(int k = min(j, sz[u]-sz[v]); k >= max(1, j-sz[v]); --k) { //枚举子树根结点所在联通块大小

f[u][j] = max(f[u][j], f[u][k]f[v][j-k]);

}

}

}

for(int i = 1; i <= sz[u]; ++i) f[u][0] = max(f[u][0], f[u][i]i);

}

int main() {

cin >> n;

for(int i = 1, x, y; i <= n-1; ++i) cin >> x >> y, addEdge(x, y), addEdge(y, x);

dp(1, 0);

cout << f[1][0] << endl;

return 0;

}

但是一交上去只有30$pts$，一看讨论区，发现还要用高精度！于是粘了个板子上去，然后就开心的$MLE$了￣▽￣。最后把$int$换成$short$就对了，无语。
粘一下$AC$代码
```cpp
#include <cstdio>
#include <iostream>
#include <cstring>

using namespace std;

#define N 700

int n, eid;
short sz[N+5], head[N+5];

struct Edge {
    int next, to;
}e[2*N+5];

struct bign{ //高精类模板，网上找的
    static const int maxn = 120;
    short d[maxn+5];
    short len;
    void clean() { while(len > 1 && !d[len-1]) len--; }
    bign() { memset(d, 0, sizeof(d)); len = 1; }
    bign(int num) { *this = num; }
    bign(char* num) { *this = num; }
    bign operator = (const char* num) {
        memset(d, 0, sizeof(d)); len = strlen(num);
        for(int i = 0; i < len; i++) d[i] = num[len-1-i] - '0';
        clean();
        return *this;
    }
    bign operator = (int num){
        char s[20]; sprintf(s, "%d", num);
        *this = s;
        return *this;
    }
    bign operator + (const bign& b) {
        bign c = *this; int i;
        for(i = 0; i < b.len; i++) {
        	c.d[i] += b.d[i];
        	if (c.d[i] > 9) c.d[i] %= 10, c.d[i+1]++;
        }
        while (c.d[i] > 9) c.d[i++] %= 10, c.d[i]++;
        c.len = max(len, b.len);
        if (c.d[i] && c.len <= i) c.len = i+1;
        return c;
    }
    bign operator - (const bign& b) {
        bign c = *this; int i;
        for(i = 0; i < b.len; i++) {
        	c.d[i] -= b.d[i];
        	if (c.d[i] < 0) c.d[i] += 10, c.d[i+1]--;
        }
        while (c.d[i] < 0) c.d[i++] += 10, c.d[i]--;
        c.clean();
        return c;
    }
    bign operator * (const bign& b) const {
        int i, j; bign c; c.len = len + b.len;
        for(j = 0; j < b.len; j++)
            for(i = 0; i < len; i++)
                c.d[i+j] += d[i]*b.d[j];
        for(i = 0; i < c.len-1; i++) c.d[i+1] += c.d[i]/10, c.d[i] %= 10;
        c.clean();
        return c;
    }
    bign operator / (const bign& b) {
    	int i, j;
        bign c = *this, a = 0;
    	for(i = len - 1; i >= 0; i--) {
    		a = a*10 + d[i];
    		for (j = 0; j < 10; j++)
                if (a < b*(j+1)) break;
    		c.d[i] = j;
    		a = a - b*j;
    	}
    	c.clean();
    	return c;
    }
    bign operator % (const bign& b) {
    	int i, j;
        bign a = 0;
    	for(i = len - 1; i >= 0; i--) {
    		a = a*10+d[i];
    		for(j = 0; j < 10; j++) if (a < b*(j+1)) break;
    		a = a-b*j;
    	}
    	return a;
    }
    bign operator += (const bign& b) {
        *this = *this+b;
        return *this;
    }
    bool operator <(const bign& b) const {
        if(len != b.len) return len < b.len;
        for(int i = len-1; i >= 0; i--)
            if(d[i] != b.d[i]) return d[i] < b.d[i];
        return false;
    }
    bool operator >(const bign& b) const { return b < *this; }
    bool operator <= (const bign& b) const { return !(b < *this); }
    bool operator >= (const bign& b) const { return !(*this < b); }
    bool operator != (const bign& b) const { return b < *this || *this < b; }
    bool operator == (const bign& b) const { return !(b < *this) && !(b > *this); }
    string str() const {
        char s[maxn] = {};
        for(int i = 0; i < len; i++) s[len-1-i] = d[i]+'0';
        return s;
    }
}f[N+5][N+5];

istream& operator >> (istream& in, bign& x) {
    string s;
    in >> s;
    x = s.c_str();
    return in;
}

ostream& operator << (ostream& out, const bign& x) {
    out << x.str();
    return out;
}

void addEdge(int u, int v) {
    e[++eid].next = head[u];
    e[eid].to = v;
    head[u] = eid;
}

void dp(int u, int fa) {
    sz[u] = 1, f[u][0] = f[u][1] = 1;
    for(int i = head[u]; i; i = e[i].next) {
        int v = e[i].to;
        if(v == fa) continue;
        dp(v, u);
        sz[u] += sz[v];
        for(int j = sz[u]; j >= 1; --j) {
            for(int k = min(j, sz[u]-sz[v]); k >= max(1, j-sz[v]); --k) {
                f[u][j] = max(f[u][j], f[u][k]*f[v][j-k]);
            }
        }
    }
    for(int i = 1; i <= sz[u]; ++i) f[u][0] = max(f[u][0], f[u][i]*i);
}

int main() {
    cin >> n;
    for(int i = 1, x, y; i <= n-1; ++i) cin >> x >> y, addEdge(x, y), addEdge(y, x);
    dp(1, 0);
    cout << f[1][0] << endl;
    return 0;
}