luogu P3232 [HNOI2013]游走
https://www.luogu.com.cn/problem/P3232
由于
m
m
m很大，所以我们先考虑经过每个点的期望次数
设
i
n
[
i
]
in[i]
in[i]表示
i
i
i点的度数，
f
i
f_i
fi表示经过
i
i
i点的期望次数
可得,假设和
i
i
i相连的点为
j
(
j
<
n
)
j(j<n)
j(j<n)
f
1
=
1
+
∑
f
j
i
n
[
j
]
\large f_1=1+\sum \frac{f_j}{in[j]}
f1=1+∑in[j]fj
f
i
=
∑
f
j
i
n
[
j
]
(
1
<
i
<
n
)
\large f_i=\sum\frac{f_j}{in[j]}(1<i<n)
fi=∑in[j]fj(1<i<n)
高斯消元可求
那么对于一条边，经过它的期望次数就是
f
u
i
n
[
u
]
+
f
v
i
n
[
v
]
\large \frac{f_u}{in[u]}+\frac{f_v}{in[v]}
in[u]fu+in[v]fv
于是乎排序贪心做即可
code:
#include<bits/stdc++.h>
#define N 505
using namespace std;
struct edge {
int v, nxt;
} e[N * N << 1];
int p[N << 1], eid;
void init() {
memset(p, -1, sizeof p);
eid = 0;
}
void insert(int u, int v) {
e[eid].v = v;
e[eid].nxt = p[u];
p[u] = eid ++;
}
double a[N][N], X[N], ans[N * N];
void guass(int n) {
for(int i = 1; i <= n; i ++) {
int j = i;
for(int k = i + 1; k <= n; k ++) if(fabs(a[k][i]) > fabs(a[j][i])) j = k;
if(i != j) swap(a[i], a[j]);
for(int k = i + 1; k <= n; k ++) {
double t = a[k][i] / a[i][i];
for(int j = i; j <= n + 1; j ++) a[k][j] -= t * a[i][j];
}
}
for(int i = n; i ; i --) {
for(int j = i + 1; j <= n; j ++) a[i][n + 1] -= a[i][j] * X[j];
X[i] = a[i][n + 1] / a[i][i];
}
}
int n, m, U[N * N], V[N * N], in[N];
int main() { init();
scanf("%d%d", &n, &m);
for(int i = 1; i <= m; i ++) {
int u, v;
scanf("%d%d", &u, &v); U[i] = u, V[i] = v;
in[u] ++, in[v] ++;
insert(u, v), insert(v, u);
}
for(int u = 1; u < n; u ++) {
a[u][u] = 1.0;
for(int i = p[u]; i + 1; i = e[i].nxt) {
int v = e[i].v;
if(v == n) continue;
a[u][v] = - 1.0 / in[v];
}
}
a[1][n] = 1.0;
// for(int i = 1; i <= n; i ++) {
// for(int j = 1; j <= n; j ++) printf("%.3lf ", a[i][j]); printf("\n");
// }
guass(n - 1);
// for(int i = 1; i <= n; i ++) printf("%.3lf ", X[i]); printf("\n");
for(int i = 1; i <= m; i ++) {
int u = U[i], v = V[i];
if(u != n) ans[i] += X[u] / in[u];
if(v != n) ans[i] += X[v] / in[v];
}
sort(ans + 1, ans + 1 + m);
double anss = 0;
for(int i = 1; i <= m; i ++) anss += ans[i] * (double)(m - i + 1);
printf("%.3lf", anss);
return 0;
}
