Codeforces 23E Tree(树型DP)
题目链接 Tree
$dp[x][i]$表示以x为根的子树中x所属的连通快大小为i的时候 答案最大值
用$dp[x][j]$ * $dp[y][k]$ 来更新$dp[x][j + k]$。
(听高手说这类题的套路其实都差不多)
因为这题输出数据会很大所以用Java……
QAQ
import java.util.*; import java.io.*; import java.math.*; public class Main{ static final int maxn = 710; static BigInteger dp[][] = new BigInteger[maxn][maxn]; static int v[][] = new int[maxn][maxn]; static int sum[] = new int[maxn]; static int n; static void dfs(int x, int pre){ int y; sum[x] = 1; for(int i = 0; i <= n; ++i) dp[x][i] = BigInteger.ONE; for(int i = 1; i <= v[x][0]; ++i){ y = v[x][i]; if(y == pre) continue; dfs(y, x); for(int j = sum[x]; j >= 0; --j) for(int k = sum[y]; k >= 0; --k){ dp[x][j + k] = dp[x][j + k].max(dp[x][j].multiply(dp[y][k])); } sum[x] += sum[y]; } for(int i = 1; i <= sum[x]; ++i) dp[x][0] = dp[x][0].max(dp[x][i].multiply(BigInteger.valueOf(i))); } public static void main(String[] args){ Scanner in = new Scanner(System.in); n = in.nextInt(); for(int i = 1; i <= n; ++i){ v[i][0] = 0; } for(int i = 1; i < n; ++i){ int x = in.nextInt(); int y = in.nextInt(); v[x][++v[x][0]] = y; v[y][++v[y][0]] = x; } dfs(1, 0); System.out.println(dp[1][0]); } }