题面
Sol
就是求期望
预处理出可可在某一位置时聪聪下一步怎么走
然后按题意模拟，记搜
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(2005);
IL ll Input(){
RG ll x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int n, m, fst[_], nxt[_], to[_], cnt, dis[_], pre[_][_], deg[_];
bool vis[_], G[_][_];
double f[_][_];
queue <int> Q;
IL void Add(RG int u, RG int v){
to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++; ++deg[v];
}
IL void Bfs(RG int S){
dis[S] = 0; Fill(vis, 0); vis[S] = 1; Q.push(S);
while(!Q.empty()){
RG int u = Q.front(); Q.pop();
for(RG int e = fst[u]; e != -1; e = nxt[e]){
if(vis[to[e]]) continue;
dis[to[e]] = dis[u] + 1;
vis[to[e]] = 1; Q.push(to[e]);
}
}
for(RG int i = 1; i <= n; ++i)
for(RG int e = fst[i]; e != -1; e = nxt[e]){
if(dis[to[e]] + 1 != dis[i]) continue;
if(!pre[S][i]) pre[S][i] = to[e];
else pre[S][i] = min(pre[S][i], to[e]);
}
}
IL double Dfs(RG int A, RG int B){
if(G[A][B]) return f[A][B];
G[A][B] = 1; RG int nt = pre[B][A];
if(A == B) return f[A][B] = 0;
if(nt == B || pre[B][nt] == B) return f[A][B] = 1;
nt = pre[B][nt]; f[A][B] = Dfs(nt, B);
for(RG int e = fst[B]; e != -1; e = nxt[e]) f[A][B] += Dfs(nt, to[e]);
f[A][B] = f[A][B] / (deg[B] + 1) + 1;
return f[A][B];
}
int main(RG int argc, RG char* argv[]){
n = Input(); m = Input(); Fill(fst, -1);
RG int A = Input(), B = Input();
for(RG int i = 1; i <= m; ++i){
RG int u = Input(), v = Input();
Add(u, v); Add(v, u);
}
for(RG int i = 1; i <= n; ++i) Bfs(i);
printf("%.3lf\n", Dfs(A, B));
return 0;
}