結果

問題 No.196 典型DP (1)
ユーザー parukiparuki
提出日時 2016-08-24 14:46:17
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 29 ms / 2,000 ms
コード長 2,855 bytes
コンパイル時間 1,771 ms
コンパイル使用メモリ 169,312 KB
実行使用メモリ 19,584 KB
最終ジャッジ日時 2024-11-08 02:03:14
合計ジャッジ時間 3,537 ms
ジャッジサーバーID
(参考情報)
judge2 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 41
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include "bits/stdc++.h"
using namespace std;
#define FOR(i,j,k) for(int (i)=(j);(i)<(int)(k);++(i))
#define rep(i,j) FOR(i,0,j)
#define each(x,y) for(auto &(x):(y))
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define debug(x) cout<<#x<<": "<<(x)<<endl
#define smax(x,y) (x)=max((x),(y))
#define smin(x,y) (x)=min((x),(y))
#define MEM(x,y) memset((x),(y),sizeof (x))
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<ll> vll;
template<int MOD>
class ModInt{
public:
ModInt():value(0){}
ModInt(long long val):value((int)(val<0?MOD+val%MOD:val%MOD)){ }
ModInt& operator+=(ModInt that){
value = value+that.value;
if(value>=MOD)value-=MOD;
return *this;
}
ModInt& operator-=(ModInt that){
value -= that.value;
if(value<0)value+=MOD;
return *this;
}
ModInt& operator*=(ModInt that){
value = (int)((long long)value * that.value % MOD);
return *this;
}
ModInt &operator/=(ModInt that){
return *this *= that.inverse();
}
ModInt operator+(ModInt that) const{
return ModInt(*this)+=that;
}
ModInt operator-(ModInt that) const{
return ModInt(*this)-=that;
}
ModInt operator*(ModInt that) const{
return ModInt(*this)*=that;
}
ModInt operator/(ModInt that) const {
return ModInt(*this) /= that;
}
ModInt pow(long long k) const{
if(value == 0)return 0;
ModInt n = *this, res = 1;
while(k){
if(k & 1)res *= n;
n *= n;
k >>= 1;
}
return res;
}
ModInt inverse() const {
long long a = value, b = MOD, u = 1, v = 0;
while(b) {
long long t = a / b;
a -= t * b;
swap(a, b);
u -= t * v;
swap(u, v);
}
return ModInt(u);
}
int toi() const{ return value; }
private:
int value;
};
typedef ModInt<1000000007> mint;
ostream& operator<<(ostream& os, const mint& x){
os << x.toi();
return os;
}
int N, K, s[2000];
vi G[2000];
mint dp[2000][2001];
void g(int u, int p){
s[u] = 1;
each(v, G[u])if(v != p){
g(v, u);
s[u] += s[v];
}
}
void f(int u, int p){
dp[u][0] = dp[u][s[u]] = 1;
int sum = 0;
each(v, G[u])if(v != p){
f(v, u);
for(int i = sum; i >= 0; --i){
FOR(j, 1, s[v] + 1){
if(i + j <= s[u]){
dp[u][i+j]+=dp[u][i]*dp[v][j];
}
}
}
sum += s[v];
}
}
int main(){
int u, v;
cin >> N >> K;
rep(i, N-1){
scanf("%d%d", &u, &v);
G[u].push_back(v);
G[v].push_back(u);
}
g(0, -1);
f(0, -1);
cout << dp[0][K] << endl;
}
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