P3232 [HNOI2013]游走
考虑求出每个边期望下经过的次数，边很多，考虑求经过每个点的个数，从而算出经过边的次数，这样排序后期望次数小的边赋值大。考虑在点上做高斯消元。因为在 $n$ 点直接停止了，所以 $n$ 不会给边带来任何贡献，所以可以不考虑 $n$ 点，以免算重。
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| #include <cstdio>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
template <class Type>
void read(Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
c == '-' && (flag = true);
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
flag && (x = ~x + 1);
}
template <class Type>
void write(Type x)
{
x < 0 && (putchar('-'), x = ~x + 1);
x > 9 && (write(x / 10), 0);
putchar(x % 10 + '0');
}
const int N = 510, M = 125e3 + 10;
int n, m, d[N];
double A[N][N], f[N];
struct Node
{
int u, v;
} e[M];
vector <int> g[N];
void Gauss ()
{
for (int k = 1; k < n; k++)
{
int t = k;
for (int i = k + 1; i < n; i++)
fabs(A[i][k]) > fabs(A[t][k]) && (t = i);
swap(A[k], A[t]);
for (int i = 1; i < n; i++)
{
if (i == k)
continue;
double t = A[i][k] / A[k][k];
for (int j = k + 1; j <= n; j++)
A[i][j] -= A[k][j] * t;
}
}
for (int i = 1; i < n; i++)
A[i][n] /= A[i][i];
}
int main ()
{
read(n), read(m);
for (int i = 1; i <= m; i++)
{
read(e[i].u), read(e[i].v);
g[e[i].u].push_back(e[i].v);
g[e[i].v].push_back(e[i].u);
d[e[i].u]++, d[e[i].v]++;
}
for (int i = 1; i < n; i++)
{
A[i][i] = 1;
for (int j : g[i])
j ^ n && (A[j][i] = -1.0 / d[i]);
}
A[1][n] = 1;
Gauss();
for (int i = 1; i <= m; i++)
f[i] = A[e[i].u][n] / d[e[i].u] + A[e[i].v][n] / d[e[i].v];
sort(f + 1, f + m + 1);
double res = 0;
for (int i = 1; i <= m; i++)
res += (m - i + 1) * f[i];
printf("%.3lf", res);
return 0;
}
|