HDU4219 Randomization?

 

HDU4219 Randomization?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 95    Accepted Submission(s): 37

Problem Description

Random is the real life. What we see and sense everyday are absolutely randomly happened. Randomization is the process of making something random, as the nature.
Given a tree with N nodes, to be more precisely, a tree is a graph in which each pair of nodes has and only has one path. All of the edges’ length is a random integer lies in interval [0, L] with equal probability. iSea want to know the probability to form a tree, whose edges’ length are randomly generated in the given interval, and in which the path's length of every two nodes does not exceed S.

 

 

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case includes three integers N, L, S. Then N - 1 lines following, each line contains two integers Ai and Bi, describing an edge of the tree.

Technical Specification
1. 1 <= T <= 512
2. 1 <= N <= 64
3. 1 <= L <= 8, 1 <= S <= 512
4. 1 <= Ai, Bi <= N

 

 

Output

For each test case, output the case number first, then the probability rounded to six fractional digits.

 

 

Sample Input

3 2 3 2 1 2 4 3 4 1 2 2 3 3 4 7 4 10 1 2 2 3 4 5 2 6 4 7 4 6

 

 

Sample Output

Case 1: 0.750000 Case 2: 0.500000 Case 3: 0.624832

 

 

Author

iSea@WHU

 

 

Source

首届华中区程序设计邀请赛暨第十届武汉大学程序设计大赛

 

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题目大意：给定一棵树，然后树的边权是[0,L]任意赋值，问这颗树不存在有一条链的长度超过S的概率。

代码：

/*

概率+树形dp

中等偏难题，想概率容易脑乱，静下心来想还是可以出的

dp[以i为根][以j为叶子节点到i的最远距离]

当j*2<=s的时候，表示这个子树上的最长链不可能超过s，那么

可以任意取值就是当前的概率，但是为了保证j是精确的，所以要

减去距离小于等于j-1的概率；

当j*2>s的时候，这个子树必定有且仅有一个链的长度是s，那么

枚举该链，让其他链的长度任意取值，所有情况之和就是概率

*/

#include <stdio.h>

#include <string.h>

#include <vector>

#define N 70

#define S 600

using namespace std;



int n,l,s,vis[N],fa[N];

vector<int>gra[N];

double dp[N][S];



void dfs(int d,int p)

{

    vis[d]=1;fa[d]=p;

    int len=gra[d].size();

    if(p!=-1&&len==1)

    {

        dp[d][0]=1;

        for(int i=1;i<=s;i++)dp[d][i]=0;

        return ;

    }

    for(int i=0;i<len;i++)

        if(!vis[gra[d][i]])dfs(gra[d][i],d);

    double sum[S]={0};

    for(int i=0;i<=s;i++)

    {

        if(i*2<=s)

        {

            dp[d][i]=1;

            for(int j=0;j<len;j++)

            {

                int e=gra[d][j];

                if(fa[e]!=d)continue;

                double tmp=0;

                for(int k=0;k<=min(l,i);k++)

                    for(int h=0;h<=i-k;h++)

                        tmp+=dp[e][h];

                tmp/=(l+1);

                dp[d][i]*=tmp;

            }

            if(i>0)dp[d][i]-=sum[i-1],sum[i]=dp[d][i]+sum[i-1];

            else sum[i]=dp[d][i];

        }

        else

        {

            int op=s-i;

            dp[d][i]=0;

            for(int j=0;j<len;j++)

            {

                int e=gra[d][j];

                if(fa[e]!=d)continue;

                double tmp1=0;

                for(int k=0;k<=min(l,op);k++)

                    for(int h=0;h<=op-k;h++)

                        tmp1+=dp[e][h];

                tmp1/=(l+1);

                double tmp2=0;

                for(int k=0;k<=min(l,i);k++)

                    tmp2+=dp[e][i-k];

                tmp2/=(l+1);

                dp[d][i]+=sum[op]*tmp2/tmp1;

            }

        }

    }

}



int main()

{

    //freopen("/home/fatedayt/in","r",stdin);

    int ncase;

    scanf("%d",&ncase);

    for(int u=1;u<=ncase;u++)

    {

        scanf("%d%d%d",&n,&l,&s);

        for(int i=1;i<=n;i++)gra[i].clear();

        for(int i=1;i<n;i++)

        {

            int a,b;

            scanf("%d%d",&a,&b);

            gra[a].push_back(b);

            gra[b].push_back(a);

        }

        memset(vis,0,sizeof(vis));

        dfs(1,-1);

        double ans=0;

        for(int i=0;i<=s;i++)ans+=dp[1][i];

        printf("Case %d: %.6lf\n",u,ans);

    }

}