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：给一张无根无向图（n个节点，m条边），每一步走哪一个儿子节点的概率相同（以哪个点为起点的概率也相同）。对于每个点，找出走完d步后走不到该点的概率。
题目分析：定义dp(s,u)表示不经过节点i(1<=i<=n)走了s步时，到达u点的概率。则根据加法原理，dp(s,son)=sum(dp(s-1,u)*f)，其中 f 是u走到其儿子son的概率，f=1.0/(u的儿子个数)。题目要求分别输出不经过节点 i (1<=i<=n) 的答案，则DP n次即可。对于每次DP，根据加法原理，答案为ans=sum(dp(d,j))，其中，j不等于i。
注意：第一步是从起点开始的，而选择起点的过程不能算做一步。
代码如下：
# include<iostream>
# include<cstdio>
# include<vector>
# include<cstring>
# include<algorithm>
using namespace std;
vector<int>v[55];
int n,m,d,vis[10005][55];
double dp[10005][55];
void solve()
{
for(int i=1;i<=n;++i){
memset(dp,0,sizeof(dp));
for(int j=1;j<=n;++j)
dp[0][j]=1.0/n;
for(int j=1;j<=d;++j){
for(int start=1;start<=n;++start){
if(i==start)
continue;
int l=v[start].size();
for(int k=0;k<l;++k)
dp[j][v[start][k]]+=dp[j-1][start]*1.0/l;
}
}
double ans=0.0;
for(int j=1;j<=n;++j)
if(j!=i)
ans+=dp[d][j];
printf("%.10lf\n",ans);
}
}
int main()
{
int T,a,b;
vector<int>::iterator it;
scanf("%d",&T);
while(T--)
{
memset(vis,0,sizeof(vis));
scanf("%d%d%d",&n,&m,&d);
for(int i=1;i<=n;++i)
v[i].clear();
while(m--)
{
scanf("%d%d",&a,&b);
it=find(v[a].begin(),v[a].end(),b);
if(it==v[a].end())
v[a].push_back(b);
it=find(v[b].begin(),v[b].end(),a);
if(it==v[b].end())
v[b].push_back(a);
}
solve();
}
return 0;
}