洛谷 P1411 树 (树形dp)

大意: 给定树, 求删除一些边, 使得连通块大小的乘积最大

 

设$dp_{i,j}$表示只考虑点$i$的子树, $i$所在连通块大小为$j$的最大值.

转移的时候不计算$i$所在连通块的贡献, 留到最后再算.

这题答案很大, 还要套个高精, 这是没套高精的代码

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 710;
int n, sz[N];
vector<int> g[N];
int dp[N][N],tmp[N];

void chkmax(int &a, int b) {a<b?a=b:0;}

void dfs(int x, int f) {
	sz[x] = dp[x][1] = 1;
	for (int y:g[x]) if (y!=f) {
		dfs(y,x);
		REP(i,1,sz[x]+sz[y]) tmp[i]=0;
		REP(i,1,sz[x]) REP(j,1,sz[y]) {
			int t = dp[x][i]*dp[y][j];
			chkmax(tmp[i],t*j);
			chkmax(tmp[i+j],t);
		}
		REP(i,1,sz[x]+sz[y]) dp[x][i]=tmp[i];
		sz[x] += sz[y];
	}
}

int main() {
	scanf("%d",&n);
	REP(i,2,n) {
		int u, v;
		scanf("%d%d",&u,&v);
		g[u].pb(v),g[v].pb(u);
	}
	dfs(1,0);
	int ans = 0;
	REP(i,1,n) ans = max(ans, dp[1][i]*i);
	printf("%d\n", ans);
}