NOI2005 : 聪聪和可可

/*
bzoj : 1415
Noi: 2005 聪聪和可可
Solution : 概率好像还是挺头疼的。
但是大多数算期望都是用dp来算的。
不会算期望只是证明你dp不好，而不是你概率不好，说白了就是自己不会。
设f[x][y]表示x - > y的期望步数，p[x][y]为x - > y的最短路中的标号最小的序号。
显然猫走两步老鼠走一步，DAG。
f[x][y] = {[\sigma dfs(p[p[x][y]][y],V)] / Size[V]} + 1;V = {v | dis[v][y] <= 1;}
翻译成人话就是x - > y的期望步数为：x向y走两步后y随机走一步(或者不走)的期望步数之和 / y的选择数 + 1;
记忆化搜索即可.
*/
#include <cmath>
#include <cstdio>
#include <cstring>
#define Rep(i,n) for(int i = 1; i <= n ; i ++)
#define RepG(i,x) for(int i = head[x] ;~ i ; i = edge[i].next)
#define CLR(a,b) memset(a,b,sizeof(a))
#define v edge[i].to
using namespace std;
const int inf = 1 << 30;
typedef long long ll;
int read(){
char ch = getchar();
while((ch < '0' || ch > '9') && ch != '-')ch = getchar ();
int x = 0;
bool flag = 0;
if(ch == '-')ch = getchar(),flag = 1;
while(ch >= '0' && ch <= '9')x = (x << 1) + (x << 3) + ch - '0',ch = getchar ();
return flag ? -x : x;
}
const int N = 1005;
double f

;
int dis

,p

;
int cnt = 0 ,head
;
void Init(){
CLR(head,-1);
cnt = 0;
}
struct Edge{int next,to;}edge[N << 1];
int n,m,S,T;
void save(int a,int b){
edge[cnt] = (Edge){head[a],b},head[a] = cnt ++;
a ^= b,b ^= a,a ^= b;
edge[cnt] = (Edge){head[a],b},head[a] = cnt ++;
}
double dfs(int x,int y)
{
if(f[x][y])return f[x][y];
if(x == y)return 0;
if(p[x][y] == y || p[p[x][y]][y] == y)return f[x][y] = 1;
double tot = dfs(p[p[x][y]][y],y),tp = 0;
RepG(i,y){
tot += dfs(p[p[x][y]][y],v);
tp ++;
}
return f[x][y] = tot / (tp + 1) + 1;
}
int q
;
void bfs(int s){
int h = 0,t = 1;
q[0] = s,dis[s][s] = 0;
while(h != t)
{
int x = q[h ++],tmp = p[s][x];
if(h == 1001)h = 0;
RepG(i,x)
if(dis[s][v] == -1 || (dis[s][x] + 1 == dis[s][v] && tmp < p[s][v]))
{
dis[s][v] = dis[s][x] + 1;
p[s][v] = tmp;
if(!tmp)p[s][v] = v;
q[t ++] = v;
if(t == 1001)t = 0;
}
}
}
int main (){
Init();
CLR(dis,-1);
n = read(),m = read();
S = read(),T = read();
Rep(i,m)
{
int a,b;
a = read(),b = read();
save(a,b);
}
Rep(i,n)bfs(i);
printf("%.3f\n",dfs(S,T));
return 0;
}