【DP】 HDOJ 5378 Leader in Tree Land

来源：互联网发布：java课程设计题目大耳编辑：程序博客网时间：2024/04/28 04:55

状态转移很好想。。。。但是复杂度并不好证明。。。但是确实是n^2的算法。。。

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int maxn = 1005;const int maxm = 20005;const int mod = 1000000007;struct Edge{int v;Edge *next;}E[maxm], *H[maxn], *edges;LL f[maxn];LL g[maxn];LL dp[maxn][maxn];LL a[maxn];int size[maxn];int tot[maxn];int n, K;LL powmod(LL a, LL b){LL base = a, res = 1;while(b) {if(b % 2) res = res * base % mod;base = base * base % mod;b /= 2;}return res;}void addedges(int u, int v){edges->v = v;edges->next = H[u];H[u] = edges++;}void init(){edges = E;memset(H, 0, sizeof H);memset(dp, 0, sizeof dp);memset(tot, 0, sizeof tot);}LL C(int n, int m){if(m > n) return 0;return f[n] * g[m] % mod * g[n - m] % mod;}void dfs(int u, int fa){dp[u][0] = 1;tot[u] = 1;size[u] = 0;for(Edge *e = H[u]; e; e = e->next) if(e->v != fa) {int v = e->v;dfs(v, u);for(int i = 0; i <= tot[u] + tot[v]; i++) a[i] = 0;LL t = C(size[u]+size[v], size[v]);for(int i = tot[u]; i >= 0; i--) {for(int j = tot[v]; j >= 1; j--) {a[i+j] = (a[i+j] + t * dp[u][i] % mod * dp[v][j] % mod) % mod;}}size[u] += size[v];tot[u] += tot[v];tot[u] = min(tot[u], K);for(int i = 0; i <= tot[u]; i++) dp[u][i] = a[i];}for(int i = 0; i <= tot[u] + 1; i++) a[i] = 0;for(int i = tot[u]; i >= 0; i--) {a[i] = (a[i] + size[u] * dp[u][i]) % mod;a[i+1] = (a[i+1] + dp[u][i]) % mod;}tot[u]++;size[u]++;for(int i = 0; i <= tot[u]; i++) dp[u][i] = a[i];}void work(){int u, v;scanf("%d%d", &n, &K);for(int i = 1; i < n; i++) {scanf("%d%d", &u, &v);addedges(u, v);addedges(v, u);}dfs(1, 1);printf("%lld\n", dp[1][K]);}int main(){f[0] = 1;for(int i = 1; i <= 1000; i++) f[i] = f[i-1] * i % mod;g[1000] = powmod(f[1000], mod - 2);for(int i = 999; i >= 0; i--) g[i] = g[i+1] * (i + 1) % mod;int _;scanf("%d", &_);for(int i = 1; i <= _; i++) {printf("Case #%d: ", i);init();work();}return 0;}

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