HDU 5001 Walk (暴力、概率dp)

Walk

Time Limit: 30000/15000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 266 Accepted Submission(s): 183 Special Judge

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

Sample Output

Source

2014 ACM/ICPC Asia Regional Anshan Online

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设dp[j][d]表示不能经过i点走了d步到达j点的概率。那么dp[j][d] = ∑ dp[k][d-1]/edge[k].size()。那么不经过i点的概率为∑dp[j][D]。（转自：http://blog.csdn.net/u013081425/article/details/39254337）（我的代码是dp[d][j]）

每次都去掉一个点求出到达其他点的概率就是不能到达这个点的概率。(转自：http://blog.csdn.net/xu12110501127/article/details/39254403)

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdlib>
 4 #include<cstdio>
 5 #include<algorithm>
 6 #include<cmath>
 7 #include<queue>
 8 #include<map>
 9 #include<string>
10 
11 #define N 55
12 #define M 15
13 #define mod 1000000007
14 #define p 10000007
15 #define mod2 100000000
16 #define ll long long
17 #define LL long long
18 #define maxi(a,b) (a)>(b)? (a) : (b)
19 #define mini(a,b) (a)<(b)? (a) : (b)
20 
21 using namespace std;
22 
23 int T;
24 int n,m,d;
25 vector<int>bian[N];
26 int cnt[N];
27 double dp[10005][N];
28 double re;
29 
30 void ini()
31 {
32     int i;
33     int x,y;
34     memset(cnt,0,sizeof(cnt));
35     scanf("%d%d%d",&n,&m,&d);
36     for(i=1;i<=n;i++){
37         bian[i].clear();
38     }
39     while(m--){
40         scanf("%d%d",&x,&y);
41         bian[x].push_back(y);
42         bian[y].push_back(x);
43     }
44     for(i=1;i<=n;i++){
45         cnt[i]=bian[i].size();
46     }
47 }
48 
49 void solve()
50 {
51     int q,o,i;
52     for(q=1;q<=n;q++)
53     {
54         re=0;
55         memset(dp,0,sizeof(dp));
56         for(i=1;i<=n;i++){
57             if(i==q) continue;
58             dp[0][i]=1.0/n;
59         }
60 
61         for(o=1;o<=d;o++){
62             for(i=1;i<=n;i++){
63                 if(i==q) continue;
64                 for(vector<int>::iterator it=bian[i].begin();it!=bian[i].end();it++){
65                     dp[o][i]+=dp[o-1][*it]/cnt[*it];
66                 }
67             }
68         }
69 
70         for(i=1;i<=n;i++){
71             if(i==p) continue;
72             re+=dp[d][i];
73         }
74         printf("%.10f
",re);
75     }
76 }
77 
78 void out()
79 {
80     //printf("%I64d
",ans);
81 }
82 
83 int main()
84 {
85     //freopen("data.in","r",stdin);
86     scanf("%d",&T);
87     //for(int cnt=1;cnt<=T;cnt++)
88     while(T--)
89     //while(scanf("%I64d%I64d",&n,&m)!=EOF)
90     {
91         ini();
92         solve();
93       //  out();
94     }
95 
96     return 0;
97 }