CF23 E. Tree 树形dp+高精度

题目链接

CF23 E. Tree

题解

CF竟让卡常QAQ

dp+高精度

dp[x][j]表示以x为根的子树,x所属的联通块大小为j,的最大乘积（不带j这块

最后f[x]维护以x为根的子树的最大答案

有点卡内存...高精压了4位

看了题解，了解到，其实这个dp的复杂度其实是O(n^2)

每次转移是复杂度是x之前的子树的sz * 当前子树的sz

相当于之前子树所有点和当前子树的点组成的点对数

而每个点对只会在lca处被计算一次

所以复杂度O(n^2)

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
inline int read() {
	int x = 0,f = 1;
	char c = getchar();
	while(c < '0' ||c > '9')c = getchar();
	while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = getchar();
	return x * f;
} 

const int L = 10000;
const int maxn = 701;
const int maxlen = 61;
struct Bignum{
	int num[maxlen];
	int len;
	Bignum () {memset(num,0,sizeof num); len = 0; }
	void print() {
		for(int i = len;i;-- i)
			if(i != len) {
				if(num[i - 1] < L / 10) {
					printf("0");
					if(num[i - 1] < L / 100) {
						printf("0");
						if(num[i - 1] < L / 1000) printf("0");
					}
				}
				printf("%d",num[i - 1]);
			} else printf("%d",num[i - 1]);
		puts("");
	}
} dp[maxn][maxn],Max[maxn]; 

Bignum operator * (Bignum a,Bignum b) {
	int len = 0;
	Bignum ret;
	for(int i = 0;i < a.len;i ++) for(int j = 0;j < b.len;j ++) {
		ret.num[i + j] += a.num[i] * b.num[j];
		if(ret.num[i + j] >= L) {
	 		ret.num[i + j + 1] += ret.num[i + j] / L;
			ret.num[i + j] %= L;
	 	}
	}
	len = a.len + b.len;
	while(ret.num[len-1] == 0 && len > 1) len --;
	ret.len = len;
	return ret;
}
Bignum operator / (Bignum a,int b) {
	int len = a.len;
	Bignum ret;
	if(!b) return ret;
	for(int i = 0;i < len;i ++) {
		ret.num[i] += a.num[i]* b;
		if(ret.num[i] >= L)  {
			ret.num[i+1] += ret.num[i] / L;
			ret.num[i] %= L;
		}
	}
	while(ret.num[len] > 0)  ret.num[len+1] = ret.num[len] / L, ret.num[len ++] %= L;
	ret.len = len;
	return ret;
}
Bignum max(Bignum a,Bignum b)  {
  	if(a.len < b.len) return b;
	if(a.len > b.len) return a;
	for(int i = a.len-1;i >= 0;i --) {
		if(a.num[i] < b.num[i]) return b;
		if(a.num[i] > b.num[i]) return a;
 	}
 	return b;
} 

//-------------------------------
struct node {
	int v,next;
}  edge[maxn << 1];
int head[maxn],num = 0;
inline void add_edge(int u,int v) {
	edge[++ num].v = v;edge[num].next = head[u];head[u] = num;
}  int n;
int siz[maxn];
void dfs(int x,int fa) {
	siz[x] = 1;
	dp[x][1].len = 1; dp[x][1].num[0] = 1;
	for(int i = head[x];i;i = edge[i].next) {
		int v = edge[i].v;
		if(v == fa) continue;
		dfs(v,x);
		siz[x] += siz[v];
	}
	for(int i = head[x];i;i = edge[i].next) {
		int v = edge[i].v;
		if(v == fa) continue;
		for(int i = siz[x];i;-- i) {
			if(dp[x][i].len)
				for(int j = 1;j <= siz[v];++ j)
					if(dp[v][j].len) dp[x][i + j] = max(dp[x][i + j],dp[x][i] * dp[v][j]);
			dp[x][i] = dp[x][i] * Max[v];
		}
	}
	for(int i = siz[x];i ;-- i) Max[x] = max(Max[x],dp[x][i] / i);
} 

int main() {
	n = read();
	for(int i = 1,u,v;i < n;++ i) {
		u = read();v = read();
		add_edge(u,v); add_edge(v,u);
	}
	dfs(1,0);
	Max[1].print();