题解来源：点击打开链接
题意:
给定n k
下面n-1行给出一棵树。
把数字1-n填到树的节点上。
填完后计算leader节点个数，若这个点是leader，则这个点上填的数>这个点的子树上填的数
问：恰好有k个leader节点的 填涂方案数.
思路:
dp[i][j]表示以i点为根的子树 有恰好j个leader的方案数。
如果u 是叶子节点则 dp[u][0] = 0, dp[u][1] = 1;
如果u不是叶子节点：
先不考虑u点能否成为leader，背包一下。
然后考虑u点：若u能成为leader,设siz[u]表示u的子树节点个数。
那么对于u的子树来说，要把[1, siz[u] ]填到子树上，当u是leader, u只能填 siz[u]
而子树的分配方案就是一个多重集的排列，因为分配给子树的是组合，子树之间是排列。
设u成为leader的方法数为 x1
x1 = (siz[u]-1)! / siz[v1]! / siz[v2]! ····
那么dp[u][i] = dp[u][i] * (u的子树填涂的总方案数 - x1) + dp[u][i-1] * x1
#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <queue>
#include <set>
#include <map>
#include <vector>
using namespace std;
template <class T>
inline bool rd(T &ret) {
char c; int sgn;
if (c = getchar(), c == EOF) return 0;
while (c != '-' && (c<'0' || c>'9')) c = getchar();
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0'&&c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return 1;
}
template <class T>
inline void pt(T x) {
if (x < 0) {
putchar('-');
x = -x;
}
if (x > 9) pt(x / 10);
putchar(x % 10 + '0');
}
typedef pair<int, int> pii;
typedef long long ll;
const int N = 1005;
const int mod = 1e9 + 7;
const int inf = 1e9;
int Pow(int x, int y) {
int ans = 1;
while (y) {
if (y & 1)ans = (ll)ans*x%mod;
y >>= 1;
x = (ll)x*x%mod;
}return ans;
}
vector<int>G[N];
int n, k;
int dp[N][N], siz[N], lef[N];
int A[N];
int mul(int x, int y) {
x = (ll)x*y%mod;
return x;
}
inline void add(int &x, int y) {
x += y; if (x >= mod)x -= mod;
}
inline void sub(int &x, int y) {
x -= y; if (x < 0)x += mod;
}
inline void dv(int &x, int y) {
x = (ll)x*Pow(y, mod - 2) % mod;
}
int g[N];
void dfs(int u, int fa) {
siz[u] = 1; lef[u] = 0;
for (auto v : G[u]) {
if (v == fa)continue;
dfs(v, u);
siz[u] += siz[v];
}
if (siz[u] == 1) {
dp[u][0] = 0; dp[u][1] = 1;
lef[u] = 1;
return;
}
dp[u][0] = 1;
int x1 = A[siz[u] - 1], x2 = A[siz[u]];
siz[u] = 0;
for (auto v : G[u]) {
if (v == fa)continue;
for (int i = lef[u] + lef[v]; i <= min(k, siz[u] + siz[v]); i++)g[i] = 0;
for (int i = lef[u]; i <= min(k, siz[u]); i++)
{
for (int j = lef[v]; j <= min(k, siz[v]) && i + j <= k; j++)
{
add(g[i + j], mul(dp[v][j], dp[u][i]));
}
}
for (int i = lef[u] + lef[v]; i <= min(k, siz[u] + siz[v]); i++)dp[u][i] = g[i];
siz[u] += siz[v];
lef[u] += lef[v];
dv(x1, A[siz[v]]);
dv(x2, A[siz[v]]);
}
siz[u]++;
sub(x2, x1);
for (int i = min(siz[u], k); i >= lef[u]; i--) {
int tmp = 0;
add(tmp, mul(dp[u][i], x2));
if (i - 1 >= lef[u])
add(tmp, mul(dp[u][i - 1], x1));
dp[u][i] = tmp;
}
}
int main() {
A[0] = 1;
for (int i = 1; i < N; i++)A[i] = (ll)A[i - 1] * i%mod;
int T, Cas = 1; rd(T);
while (T--) {
rd(n); rd(k);
for (int i = 1; i <= n; i++)G[i].clear(), memset(dp[i], 0, sizeof dp[i]);
for (int i = 1, u, v; i < n; i++) {
rd(u); rd(v);
G[u].push_back(v); G[v].push_back(u);
}
dfs(1, 1);
printf("Case #%d: ", Cas++);
pt(dp[1][k]); puts("");
}
return 0;
}
/*
99
5 3
1 2
2 3
2 4
1 5
4 2
1 2
2 3
2 4
ans:12
4 3
1 2
2 3
2 4
4 4
1 2
2 3
2 4
*/
Total Submission(s): 262 Accepted Submission(s): 88
Problem Description
Tree land has
n
cities, connected by
n−1
roads. You can go to any city from any city. In other words, this land is a tree. The city numbered one is the root of this tree.
There are n ministers numbered from 1 to n . You will send them to n cities, one city with one minister.
Since this is a rooted tree, each city is a root of a subtree and there are n subtrees. The leader of a subtree is the minister with maximal number in this subtree. As you can see, one minister can be the leader of several subtrees.
One day all the leaders attend a meet, you find that there are exactly k ministers. You want to know how many ways to send n ministers to each city so that there are k ministers attend the meet.
Give your answer mod 1000000007 .
There are n ministers numbered from 1 to n . You will send them to n cities, one city with one minister.
Since this is a rooted tree, each city is a root of a subtree and there are n subtrees. The leader of a subtree is the minister with maximal number in this subtree. As you can see, one minister can be the leader of several subtrees.
One day all the leaders attend a meet, you find that there are exactly k ministers. You want to know how many ways to send n ministers to each city so that there are k ministers attend the meet.
Give your answer mod 1000000007 .
Input
Multiple test cases. In the first line there is an integer
T
, indicating the number of test cases. For each test case, first line contains two numbers
n,k
. Next
n−1
line describe the roads of tree land.
T=10,1≤n≤1000,1≤k≤n
T=10,1≤n≤1000,1≤k≤n
Output
For each test case, output one line. The output format is Case #
x
:
ans
,
x
is the case number,starting from
1
.
Sample Input
2
3 2
1 2
1 3
10 8
2 1
3 2
4 1
5 3
6 1
7 3
8 7
9 7
10 6
Sample Output
Case #1: 4
Case #2: 316512
Source
