【HNOI2013】游走

题面

题解

图上的期望大部分是\(dp\)，无向图的期望大部分是高斯消元

设\(f[i]\)表示走到点\(i\)的期望，\(d[i]\)表示\(i\)的度，\(to(i)\)表示\(i\)能到达的点集

所以\(f[i] = \sum\limits_{x \in to(i)} f[x] / d[x]\)

然后每个点能够列出这样的方程，直接高斯消元就可以了

代码

#include<bits/stdc++.h>
#define RG register
#define clear(x, y) memset(x, y, sizeof(x));
using namespace std;

inline int read()
{
	int data = 0, w = 1;
	char ch = getchar();
	while(ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if(ch == '-') w = -1, ch = getchar();
	while(ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
	return data*w;
}

const int maxn(510), maxm(250100);
struct edge { int next, to; } e[maxm << 1];
int head[maxn], e_num;
inline void add_edge(int from, int to) { e[++e_num] = {head[from], to}; head[from] = e_num; }
double a[maxn][maxn], ans[maxm], Ans, deg[maxn];
int n, m, from[maxm], to[maxm];

inline void Gauss()
{
	for(RG int i = 1, k = i; i <= n; i++, k = i)
	{
		for(RG int j = k + 1; j <= n; j++) if(fabs(a[k][i]) < fabs(a[j][i])) k = j;
		swap(a[i], a[k]);
		for(RG int j = i + 1; j <= n + 1; j++) a[i][j] /= a[i][i];
		a[i][i] = 1.;
		for(RG int j = 1; j <= n; j++)
		{
			if(i == j) continue;
			for(RG int k = i + 1; k <= n + 1; k++) a[j][k] -= a[j][i] * a[i][k];
			a[j][i] = 0.;
		}
	}
}

int main()
{
	n = read(); m = read();
	for(RG int i = 1; i <= m; i++)
	{
		from[i] = read(); to[i] = read();
		add_edge(from[i], to[i]); deg[from[i]] += 1.;
		add_edge(to[i], from[i]); deg[to[i]] += 1.;
	}
	for(RG int i = 1; i < n; i++)
	{
		for(RG int j = head[i]; j; j = e[j].next) if(e[j].to != n) a[i][e[j].to] += -1. / deg[e[j].to];
		a[i][i] = 1;
	}
	a[n][n] = 1;
	a[1][n + 1] = 1; Gauss();
	for(RG int i = 1; i <= m; i++)
		ans[i] = ((from[i] == n) ? 0 : a[from[i]][n + 1] / deg[from[i]]) + ((to[i] == n) ? 0 : a[to[i]][n + 1] / deg[to[i]]);
	sort(ans + 1, ans + m + 1);
	for(RG int i = 1; i <= m; i++) Ans += (m - i + 1) * ans[i];
	printf("%.3lf\n", Ans);
	return 0;
}