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.
题意：有一张双连通图，有个人在图上等概率随机选择起点，每一步等概率随机选择下一个走的点，一共走 d 步，问每个点没有被走到的概率是多少。
组数20，点数50，边数2450，总步数10000，暴力更新复杂度 20 × 50 × 2450 × 10000，重点：时限15秒！
随便暴……
dp [ i ] [ j ] [ k ] 表示当前第 i 轮，正位于 j 点，途中没有经过 k 点的概率总和。
每一轮在 j 点确定下一个点走哪个点时，其概率都可以累加到没有走到其他点的概率上。
数组开不下用滚动。
1 #include<stdio.h>
2 #include<string.h>
3
4 int head[55],nxt[2550],point[2550],size;
5 int num[55];
6 double a[2][55][55],ans[55];
7 //int ma[55][55]
8
9 void add(int a,int b){
10 point[size]=b;
11 nxt[size]=head[a];
12 head[a]=size++;
13 num[a]++;
14 point[size]=a;
15 nxt[size]=head[b];
16 head[b]=size++;
17 num[b]++;
18 }
19
20 int main(){
21 int T;
22 scanf("%d",&T);
23 while(T--){
24 int n,m,d;
25 scanf("%d%d%d",&n,&m,&d);
26 memset(head,-1,sizeof(head));
27 size=0;
28 memset(num,0,sizeof(num));
29 while(m--){
30 int u,v;
31 scanf("%d%d",&u,&v);
32 add(u,v);
33 // ma[u][v]=ma[v][u]=1;
34 }
35 for(int i=1;i<=n;++i){
36 for(int j=1;j<=n;++j){
37 if(i!=j)a[0][i][j]=1.0/n;
38 else a[0][i][j]=0;
39 }
40 }
41 int o=0;
42 // if(d>1000)d=1000;
43 for(int i=1;i<=d;++i){
44 memset(a[o^1],0,sizeof(a[o^1]));
45 for(int j=1;j<=n;++j){
46 for(int u=head[j];~u;u=nxt[u]){
47 int v=point[u];
48 for(int k=1;k<=n;++k){
49 if(j!=k)a[o^1][j][k]+=a[o][v][k]/num[v];
50 }
51 }
52 }
53 o^=1;
54 }
55 for(int i=1;i<=n;++i){
56 ans[i]=0;
57 for(int j=1;j<=n;++j)ans[i]+=a[o][j][i];
58 }
59 for(int i=1;i<=n;++i)printf("%.10lf
",ans[i]);
60 }
61 return 0;
62 }