hdu5001 Walk 概率DP

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。

 #include<stdio.h>
 #include<string.h>

 int head[],nxt[],point[],size;
 int num[];
 double a[][][],ans[];
 //int ma[55][55]

 void add(int a,int b){
     point[size]=b;
     nxt[size]=head[a];
     head[a]=size++;
     num[a]++;
     point[size]=a;
     nxt[size]=head[b];
     head[b]=size++;
     num[b]++;
 }

 int main(){
     int T;
     scanf("%d",&T);
     while(T--){
         int n,m,d;
         scanf("%d%d%d",&n,&m,&d);
         memset(head,-,sizeof(head));
         size=;
         memset(num,,sizeof(num));
         while(m--){
             int u,v;
             scanf("%d%d",&u,&v);
             add(u,v);
         //    ma[u][v]=ma[v][u]=1;
         }
         for(int i=;i<=n;++i){
             for(int j=;j<=n;++j){
                 if(i!=j)a[][i][j]=1.0/n;
                 else a[][i][j]=;
             }
         }
         int o=;
     //    if(d>1000)d=1000;
         for(int i=;i<=d;++i){
             memset(a[o^],,sizeof(a[o^]));
             for(int j=;j<=n;++j){
                 for(int u=head[j];~u;u=nxt[u]){
                     int v=point[u];
                     for(int k=;k<=n;++k){
                         if(j!=k)a[o^][j][k]+=a[o][v][k]/num[v];
                     }
                 }
             }
             o^=;
         }
         for(int i=;i<=n;++i){
             ans[i]=;
             for(int j=;j<=n;++j)ans[i]+=a[o][j][i];
         }
         for(int i=;i<=n;++i)printf("%.10lf\n",ans[i]);
     }
     return ;
 }