HDU 5001 Walk (概率dp)

原题链接

Problem Description

I used to think I could be anything, but now I know that I couldn’t do anything. So I started traveling.

The nation looks like a connected bidirectional graph, and I am randomly walking on it. It means when I am at node i, I will travel to an adjacent node with the same probability in the next step. I will pick up the start node randomly (each node in the graph has the same probability.), and travel for d steps, noting that I may go through some nodes multiple times.

If I miss some sights at a node, it will make me unhappy. So I wonder for each node, what is the probability that my path doesn’t contain it.

Input

The first line contains an integer T, denoting the number of the test cases.

For each test case, the first line contains 3 integers n, m and d, denoting the number of vertices, the number of edges and the number of steps respectively. Then m lines follows, each containing two integers a and b, denoting there is an edge between node a and node b.

T<=20, n<=50, n-1<=m<=n*(n-1)/2, 1<=d<=10000. There is no self-loops or multiple edges in the graph, and the graph is connected. The nodes are indexed from 1.

Output

For each test cases, output n lines, the i-th line containing the desired probability for the i-th node.

Your answer will be accepted if its absolute error doesn’t exceed 1e-5.

Sample Input

2
5 10 100
1 2
2 3
3 4
4 5
1 5
2 4
3 5
2 5
1 4
1 3
10 10 10
1 2
2 3
3 4
4 5
5 6
6 7
7 8
8 9
9 10
4 9

Sample Output

0.0000000000
0.0000000000
0.0000000000
0.0000000000
0.0000000000
0.6993317967
0.5864284952
0.4440860821
0.2275896991
0.4294074591
0.4851048742
0.4896018842
0.4525044250
0.3406567483
0.6421630037

题目大意

在图上随机选一个点，然后开始一相同的概率往下走，问d步之后每个点没被走到的概率。

解题思路

用dp [i] [j] 表示走了i步之后到达j的概率，所以有转移方程dp[j][k]+=dp[j-1][map[k][l]]/map[k].size(); 统计最后一步到达其他点的概率之和即为不能到达当前点的概率。

AC代码

#include<bits/stdc++.h>#define ll long long#define rep(i,a,b) for (int i=(a),_ed=(b);i<=_ed;i++)#define fil(a,b) memset((a),(b),sizeof(a))#define cl(a) fil(a,0)#define PI 3.1415927#define inf 0x3f3f3f3fusing namespace std;double dp[10005][55];vector<int> map[55];int main(){    int t,n,m,d,from,to;    cin>>t;    while(t--)    {        cin>>n>>m>>d;        for(int i=0; i<55; i++) map[i].clear();        rep(i,1,m)        {            scanf("%d%d",&from,&to);            map[from].push_back(to);            map[to].push_back(from);        }        rep(i,1,n) dp[0][i]=1.0/n;        rep(i,1,n)        {            rep(j,1,d)            {                rep(k,1,n)                {                    if(i!=k)                    {                        dp[j][k]=0;                        for(int l=0;l<map[k].size();l++)                        {                            if(map[k][l]!=i) dp[j][k]+=dp[j-1][map[k][l]]/map[k].size();                        }                    }                }            }            double ans=0;            rep(j,1,n)            {                if(j!=i)                {                    ans+=dp[d][j];                }            }            printf("%.6lf\n",ans);        }    }    return 0;}