HDU - 5378 Leader in Tree Land 树形dp

HDU - 5378

dp[ i ][ j ]表示 i 这棵子树填满， 有 j 个满足条件的方案数。

转移的时候组合数乘一乘。

#include
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair
#define PLI pair
#define PII pair
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1000 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n, k, C[N][N];
int dp[N][N];
int tmp[N];
int sz[N];
vector<int> G[N];


void dfs(int u, int fa) {
    sz[u] = 1;
    int comb = 1;
    for(auto &v : G[u]) {
        if(v == fa) continue;
        dfs(v, u);
        sz[u] += sz[v];
    }
    int nowsz = sz[u] - 1;
    for(auto &v : G[u]) {
        if(v == fa) continue;
        comb = 1LL * comb * C[nowsz][sz[v]] % mod;
        nowsz -= sz[v];
    }

    memset(tmp, 0, sizeof(tmp));
    tmp[0] = 1;
    nowsz = 0;
    for(auto &v : G[u]) {
        if(v == fa) continue;
        for(int i = nowsz; i >= 0; i--) {
            for(int j = 1; j <= sz[v]; j++) {
                if(i + j > k) break;
                add(tmp[i + j], 1LL * tmp[i] * dp[v][j] % mod);
            }
            tmp[i] = 0;
        }
        nowsz += sz[v];
    }
    for(int i = 1; i <= min(sz[u], k); i++) {
        dp[u][i] = 1LL * tmp[i] * comb % mod * (sz[u] - 1) % mod;
    }
    for(int i = min(sz[u], k) - 1; i >= 0; i--) {
        add(dp[u][i + 1], 1LL * tmp[i] * comb % mod);
    }
}

void init() {
    for(int i = 1; i <= n; i++) {
        G[i].clear();
    }
}

int main() {
    for(int i = 0; i < N; i++) {
        for(int j = C[i][0] = 1; j <= i; j++) {
            C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
        }
    }
    int T; scanf("%d", &T);
    for(int cas = 1; cas <= T; cas++) {
        scanf("%d%d", &n, &k);
        init();
        for(int i = 1; i < n; i++) {
            int u, v;
            scanf("%d%d", &u, &v);
            G[u].push_back(v);
            G[v].push_back(u);
        }
        dfs(1, 0);
        printf("Case #%d: %d\n", cas, dp[1][k]);
    }
    return 0;
}

/*
*/

 

转载于:https://www.cnblogs.com/CJLHY/p/11195252.html