Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P2796 Facer的程序

P2796 Facer的程序

比较简单的树形DP,求树上子树个数。

对于一个节点 $x$ ,在每一个子节点 $y$ 都有 $f[y] + 1$ 种选法。

查看代码
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#include <cstdio>
using namespace std;
typedef long long LL;
const int N = 1e5 + 10, M = 2e5 + 10, mod = 1e9 + 7;
int n;
LL f[N];
int idx, hd[N], nxt[M], edg[M];
void dfs(int x, int fa)
{
f[x] = 1;
for (int i = hd[x]; ~i; i = nxt[i])
if (edg[i] != fa)
{
dfs(edg[i], x);
f[x] = f[x] * (f[edg[i]] + 1) % mod;
}
}
void add(int a, int b)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
}
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++)
hd[i] = -1;
for (int i = 1, a, b; i < n; i++)
{
scanf("%d%d", &a, &b);
add(a, b);
add(b, a);
}
dfs(1, 0);
LL res = 0;
for (int i = 1; i <= n; i++)
res = (res + f[i]) % mod;
printf("%lld", res);
return 0;
}