题面
Sol
(Zsy)又在切大火题了
考虑暴力，因为它是无穷的，我们可以设(f[i][j])表示走了(i)条边，到达(j)的概率，然后跑(5000)步就有(50)分
那么边经过次数的期望就可以算出来
对这些边的期望排序，一一编号，注意(n)不要转移
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(505);
const int __(500005);
IL ll Input(){
RG char c = getchar(); RG ll x = 0, z = 1;
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[_], nt[__], cnt, to[__], id[__], dg[_];
double f[2][_], g[__], ans;
IL void Add(RG int u, RG int v, RG int ID){
id[cnt] = ID; to[cnt] = v; nt[cnt] = fst[u]; fst[u] = cnt++; ++dg[v];
}
int main(RG int argc, RG char* argv[]){
n = Input(); m = Input(); Fill(fst, -1);
for(RG int i = 1, a, b; i <= m; ++i)
a = Input(), b = Input(), Add(a, b, i), Add(b, a, i);
f[0][1] = 1;
for(RG int d = 1, lst = 0, nxt = 1; d <= 5000; ++d){
for(RG int u = 1; u < n; ++u){
if(f[lst][u] == 0) continue;
for(RG int e = fst[u]; e != -1; e = nt[e]){
f[nxt][to[e]] += f[lst][u] / dg[u];
g[id[e]] += f[lst][u] / dg[u];
}
f[lst][u] = 0;
}
swap(lst, nxt);
}
sort(g + 1, g + m + 1);
for(RG int i = 1; i <= m; ++i) ans += g[i] * (m - i + 1);
printf("%.3lf
", ans);
return 0;
}
考虑优化
换一种思维，设每条边经过次数的期望为(g[i])，每个点的度数为(dg[i])，到每个点的概率为(f[i])
设第(i)条边连接(u,v)
那么(g[i] = frac{f[u]}{dg[u]}+frac{f[v]}{dg[v]})
设(i)连接的点集为(S)
那么(f[i] = sum_{jin S}frac{f[j]}{dg[j]})
那么就可以解方程组求出所有的(f[i])，从而(g)也就求出来了
注意(f[1])最开始为(1)，注意(n)不要转移
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(505);
const int __(500005);
IL ll Input(){
RG char c = getchar(); RG ll x = 0, z = 1;
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[__], cnt, to[__], dg[_], ff[__], tt[__];
double g[__], ans, f[_], a[_][_];
IL void Add(RG int u, RG int v){
to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++; ++dg[v];
}
IL void Gauss(){
for(RG int i = 1; i < n; ++i)
for(RG int j = i + 1; j <= n; ++j){
RG double div = a[j][i] / a[i][i];
for(RG int k = 1; k <= n + 1; ++k) a[j][k] -= a[i][k] * div;
}
for(RG int i = n; i; --i){
f[i] = a[i][n + 1] / a[i][i];
for(RG int j = i - 1; j; --j) a[j][n + 1] -= f[i] * a[j][i];
}
}
int main(RG int argc, RG char* argv[]){
n = Input(); m = Input(); Fill(fst, -1);
for(RG int i = 1; i <= m; ++i)
ff[i] = Input(), tt[i] = Input(), Add(ff[i], tt[i]), Add(tt[i], ff[i]);
for(RG int u = 1; u < n; ++u){
a[u][u] = -1;
for(RG int e = fst[u]; e != -1; e = nxt[e]) a[to[e]][u] = 1.0 / dg[u];
}
a[n][n] = a[1][n + 1] = -1; Gauss();
for(RG int i = 1; i <= m; ++i){
if(ff[i] != n) g[i] = f[ff[i]] / dg[ff[i]];
if(tt[i] != n) g[i] += f[tt[i]] / dg[tt[i]];
}
sort(g + 1, g + m + 1);
for(RG int i = 1; i <= m; ++i) ans += g[i] * (m - i + 1);
printf("%.3lf
", ans);
return 0;
}