2015 Multi-University Training Contest 7 hdu 5378 Leader in Tree Land

Leader in Tree Land

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 447    Accepted Submission(s): 185

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 $1000000007$.

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.

$T=10,1 \leq n \leq 1000,1 \leq k \leq n$

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

Author
UESTC

Source
2015 Multi-University Training Contest 7  1010

解题：动态规划 这位博主大牛说得非常清楚

我们用x[i],y[i]分别代表这个节点能够成为leader和不能够成为leader的概率。

son[i] 代表以i节点为根的子树的节点数。

那么$x[i] = 1/son[i]，y[i] = 1-(1/son[i])$。因为这里面出现了分数，所有我们用逆元处理一下。

我们设dp[i][j]表示编号为1,2...i的节点中有j个leader的概率。

那么转移方程就是 $dp[i][j] = dp[i-1][j-1] * x[i] + dp[i-1][j] * y[i]$。

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const LL mod = ;
 const int maxn = ;
 struct arc {
     int to,next;
     arc(int x = ,int y = -) {
         to = x;
         next = y;
     }
 } e[maxn<<];
 int son[maxn],head[maxn],tot;
 LL F[maxn] = {},x[maxn],y[maxn],dp[maxn][maxn];
 void init() {
     for(int i = ; i < maxn; ++i)
         F[i] = F[i-]*i%mod;
 }
 void add(int u,int v) {
     e[tot] = arc(v,head[u]);
     head[u] = tot++;
 }
 LL gcd(LL a,LL b,LL &x,LL &y) { //ax + by = gcd(a,b)
     if(!b) {
         x = ;
         y = ;
         return a;
     }
     LL ret = gcd(b,a%b,y,x);
     y -= x*(a/b);
     return ret;
 }
 LL Inv(LL b,LL mod) { //求b % mod的逆元
     LL x,y,d = gcd(b,mod,x,y);
     return d == ?(x%mod + mod)%mod:-;
 }
 void dfs(int u,int fa) {
     son[u] = ;
     for(int i = head[u]; ~i; i = e[i].next) {
         if(e[i].to == fa) continue;
         dfs(e[i].to,u);
         son[u] += son[e[i].to];
     }
 }
 int main() {
     int kase,n,k,u,v,cs = ;
     init();
     scanf("%d",&kase);
     while(kase--) {
         scanf("%d%d",&n,&k);
         memset(head,-,sizeof head);
         tot = ;
         for(int i = ; i < n; ++i) {
             scanf("%d%d",&u,&v);
             add(u,v);
             add(v,u);
         }
         dfs(,-);
         for(int i = ; i <= n; ++i) {
             x[i] = Inv(son[i],mod);
             y[i] = (son[i] - )*x[i]%mod;
         }
         memset(dp,,sizeof dp);
         dp[][] = y[];
         dp[][] = x[];
         for(int i = ; i <= n; ++i) {
             dp[i][] = dp[i-][]*y[i]%mod;
             for(int j = ; j <= min(i,k); ++j) {
                 LL a = dp[i-][j-]*x[i]%mod;
                 LL b = dp[i-][j]*y[i]%mod;
                 dp[i][j] = (a + b)%mod;
             }
         }
         printf("Case #%d: %I64d\n",cs++,dp[n][k]*F[n]%mod);
     }
     return ;
 }