ARC101E - Ribbons on Tree

题目链接

ARC101E - Ribbons on Tree

题解

令边集\(S \subseteq E\) 设\(f(S)\)为边集S中没有边被染色的方案数

容斥一下，那么\(ans = \sum_{S \subseteq E} (-1)^{ \| S\| f(S) }\)

那么如何求对于原边集的\(f(S)\),也就是把\(S\)集合中的边全部删掉之后的各联通块内匹配的乘积

设\(g(x)\)为大小为x的联通块内点两两匹配的方案

那么\(f(S)=\prod_{i=1}^{|S|+1}g(a_i)\)

考虑如何求ans

设\(dp[i][j]\)表示以i为跟的子树中，有j各点没有在子树种匹配(链接到父节点

转移背包一下

对于\(j=0\)的时候由于那么i节点到父亲的边是没有覆盖的，容斥系数要取反

那么

$ f[i][0]=\sum_{j=1}^{sz[i]}-1\times f[i][j]\times g(j) $

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
#define rep(p,x,k) for(int p = x;p <= k;++ p)
#define per(p,x,k) for(int p = x;p >= k;-- p)
#define gc getchar()
#define pc putchar
#define LL long long
#define int long long
inline LL read() {
    LL x = 0,f = 1;
    char c = gc;
    while(c < '0' || c > '9') c = gc;
    while(c <= '9' && c >= '0') x = x * 10 + c -'0',c = gc;
    return x ;
}
void print(LL x) {
    if(x < 0) {
        pc('-');
        x = -x;
    }
    if(x >= 10) print(x / 10);
    pc(x % 10 + '0');
}
const int maxn = 5007;
const int mod = 1e9 + 7;
int a[maxn];
int n;
struct node {
	int v,next;
} edge[maxn << 1];
int num = 0,head[maxn];
inline void add_edge(int u,int v) {
	edge[++ num].v = v; edge[num].next = head[u];head[u] = num;
}
int g[maxn];
int dp[maxn][maxn];
int siz[maxn];
void dfs(int x,int fa) {
	static int t[maxn];
	dp[x][1] = 1; siz[x] = 1;
	for(int i = head[x];i;i = edge[i].next) {
		int v = edge[i].v;
		if(v == fa) continue;
		dfs(v,x);
		for(int j = 0,kel = siz[v];j <= siz[x];++ j) {
			for(int k = 0;k <= siz[v];++ k) {
				(t[j + k] += 1ll * dp[x][j] * dp[v][k] % mod) %= mod;
			}
		}
		siz[x] += siz[v];
		for(int j = 0;j <= siz[x];++ j) dp[x][j] = t[j],t[j] = 0;
	}
	LL sum = 0;
	for(int i = 0;i <= siz[x];i += 2) sum += mod - 1ll * dp[x][i] * g[i] % mod;
	dp[x][0] = sum % mod;
}
 main() {
	n = read();
	int u,v;
	rep(i, 1,n - 1) {
		u = read(),v = read();
		add_edge(u,v);
		add_edge(v,u);
	}
	g[0] = 1;
	for(int i = 2;i <= n;i += 2) (g[i] = 1ll * g[i - 2] * (i - 1)) %= mod;
	dfs(1,1);
	print((mod - dp[1][0]) % mod);
	return 0;
}
/*
4
1 2
1 3
1 4
*/