P4206 聪聪和可可

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聪聪和可可

鸽了两天

设$i$,$j$之间有一条边则 $$ if\left(dis[i][k]-1==dis[k][j]\right) \\ next[i][k]=\min(next[i][k],j) $$ 考虑答案的转移，设$u$为$cat$当前所在的位置，$v$为$mouse$当前所在的位置，$a$为$cat$走一步所到的位置，$b$为$cat$走两步所到的位置，$to$为$mouse$可到达的位置~~最后记得算上一开始的v~~，$du$数组为该点的出度~~记得加上自己~~，则有以下三种情况 $$ if\left(u==v\right)\;\;return 0; \\ if(a==v||b==v)\;\;return 1; \\ else\;\;dp[u][v]+=\frac{dfs\left(b,to\right)}{\left(du[v]+1\right)} $$ 最后记得算上$mouse$一开始所在的位置$v dp[u][v]+=\frac{dfs\left(b,v\right)}{\left(du[v]+1\right)}

以上求答案的过程可用dfs解决

Code

#include<cstdio>
#include<queue>
#define MAX 1001
#define re register
#define INF 114514
namespace OMA
{
   int n,e,c,m;
   int dis[MAX][MAX];
   int cnt=1,head[MAX];
   int du[MAX],next[MAX][MAX];
   double dp[MAX][MAX];
   bool pre[MAX],vis[MAX][MAX];
   struct node
   {
     int next;
     int to;
   }edge[MAX<<1];
   typedef std::pair<int,int>oma;
   std::priority_queue<oma,std::vector<oma>,std::greater<oma> >q;
   inline int read()
   {
     int s=0,w=1; char ch=getchar();
     while(ch<'0'||ch>'9'){ if(ch=='-')w=-1; ch=getchar(); }
     while(ch>='0'&&ch<='9'){ s=s*10+ch-'0'; ch=getchar(); }
     return s*w;
   }
   inline void add(int u,int v)
   {
     edge[++cnt].next = head[u];
     edge[cnt].to = v;
     head[u] = cnt;
   }
   inline int min(int a,int b)
   { return a<b?a:b; }
   void in()
   {
     n=read(),e=read(),c=read(),m=read();
     for(re int i=1; i<=e; i++)
     {
       int u=read(),v=read();
       add(u,v),add(v,u);
       du[u]++,du[v]++;
     }
     for(re int i=1; i<=n; i++)
     {
       for(re int j=1; j<=n; j++)
       { next[i][j] = INF; }
     }
   }
   inline void dijkstra(int u)
   {
     for(re int i=1; i<=n; i++)
     { dis[u][i] = INF ,pre[i] = false; }
     while(!q.empty())
     { q.pop(); }
     dis[u][u] = 0;
     q.push(std::make_pair(0,u));
     while(!q.empty())
     {
       int k = q.top().second;
       q.pop();
       if(pre[k])
       { continue; }
       pre[k] = true;
       for(re int i=head[k]; i; i=edge[i].next)
       {
         int to = edge[i].to;
         if(!pre[to]&&(dis[u][to]>dis[u][k]+1))
         {
           dis[u][to] = dis[u][k]+1;
           q.push(std::make_pair(dis[u][to],to));
         }
       }
     }
   }
   inline double dfs(int cat,int mouse)
   {
     if(vis[cat][mouse])
     { return dp[cat][mouse]; }
     if(cat==mouse)
     { return 0; }
     int a,b;
     a = next[cat][mouse];
     b = next[a][mouse];
     if(a==mouse||b==mouse)
     { return 1; }
     dp[cat][mouse] = 1;
     for(re int i=head[mouse]; i; i=edge[i].next)
     {
       int to = edge[i].to;
       dp[cat][mouse] += dfs(b,to)/(du[mouse]+1);
     }
     dp[cat][mouse] += dfs(b,mouse)/(du[mouse]+1);
     vis[cat][mouse] = 1;
     return dp[cat][mouse];
   }
   signed main()
   {
     in();
     for(re int i=1; i<=n; i++)
     { dijkstra(i); }
     for(re int i=1; i<=n; i++)
     {
       for(re int j=head[i]; j; j=edge[j].next)
       {
         int to = edge[j].to;
         for(re int k=1; k<=n; k++)
         {
           if(dis[i][k]-1==dis[k][to])
           { next[i][k] = min(next[i][k],to); }
         }
       }
     }
     printf("%.3lf\n",dfs(c,m));
     return 0;
   }
}
signed main()
{ return OMA::main(); }