HDOJ 5001 Walk

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機率DP

dp[j][d] 表示不經過i點走d步到j的機率, dp[j][d]=sigma ( dp[k][d-1] * Probability )

ans = sigma ( dp[j][D] )

WalkTime Limit: 30000/15000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 401    Accepted Submission(s): 261
Special Judge


Problem DescriptionI 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. 
InputThe 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. 
OutputFor 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
25 10 1001 22 33 44 51 52 43 52 51 41 310 10 101 22 33 44 55 66 77 88 99 104 9
 
Sample Output
0.00000000000.00000000000.00000000000.00000000000.00000000000.69933179670.58642849520.44408608210.22758969910.42940745910.48510487420.48960188420.45250442500.34065674830.6421630037
 
Source2014 ACM/ICPC Asia Regional Anshan Online 


#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;const int maxn=10010;int n,m,D;vector<int> g[maxn];double dp[55][maxn];int main(){int T_T;scanf("%d",&T_T);while(T_T--){scanf("%d%d%d",&n,&m,&D);for(int i=0;i<=n+1;i++) g[i].clear();while(m--){int a,b;scanf("%d%d",&a,&b);g[a].push_back(b);g[b].push_back(a);}for(int i=1;i<=n;i++){memset(dp,0,sizeof(dp));for(int j=1;j<=n;j++){if(i!=j) dp[j][0]=1.0/n;}for(int d=1;d<=D;d++){for(int j=1;j<=n;j++){if(j==i) continue;for(int k=0,sz=g[j].size();k<sz;k++){int v=g[j][k];if(v!=i) dp[j][d]+=dp[v][d-1]*(1./g[v].size());}}}double ans=0.0;for(int j=1;j<=n;j++){if(i!=j) ans+=dp[j][D];}printf("%.10lf\n",ans);}}return 0;}



HDOJ 5001 Walk

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