BZOJ2337 [HNOI2011]XOR和路径 【概率dp + 高斯消元】

题目

题解

突然get到这样路径期望的题目八成是高斯消元

因为路径上的dp往往具有后效性，这就形成了一个方程组

对于本题来说，直接对权值dp很难找到突破口

但是由于异或是位独立的，我们考虑求出每一位的期望

设\(f[i]\)为从节点\(i\)出发到达N的期望值

有\(f[i] = \frac{f[j]}{degree[i]} + \frac{1 - f[k]}{degree[i]} [edge(i,j) = 0,edge(i,k) = 1]\)

因为如果出边权值为0，异或之后值不变，等于\(f[j]\)的值，

如果权值为1，异或后取反，等于\(1-f[k]\)

同时\(f[n] = 0\)

列出式子后就是一个n元方程组

最后要注意自环只算该点的一条边

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#define eps 1e-9
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u]; k; k = ed[k].nxt)
using namespace std;
const int maxn = 105,maxm = 100005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - 48; c = getchar();}
	return out * flag;
}
int h[maxn],ne = 2;
double de[maxn];
struct EDGE{int to,nxt,w;}ed[maxm];
inline void build(int u,int v,int w){
	ed[ne] = (EDGE){v,h[u],w}; h[u] = ne++;
	if (u != v){
		ed[ne] = (EDGE){u,h[v],w}; h[v] = ne++;
		de[u] += 1,de[v] += 1;
	}
	else de[u] += 1;
}
int n,m,p;
double A[maxn][maxn],ans;
void gause(){
	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 (fabs(A[j][i]) < eps) exit(0);
		double t = A[j][i];
		for (int k = i; k <= n + 1; k++) swap(A[i][k],A[j][k]),A[i][k] /= t;
		for (j = i + 1; j <= n; j++){
			if (fabs(A[j][i]) > eps){
				t = A[j][i];
				for (int k = i; k <= n + 1; k++)
					A[j][k] -= A[i][k] * t;
			}
		}
	}
	for (int i = n; i; i--){
		for (int j = n; j > i; j--)
			A[i][n + 1] -= A[i][j] * A[j][n + 1];
		A[i][n + 1] /= A[i][i];
	}
}
void solve(){
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n + 1; j++)
			A[i][j] = 0;
	for (int i = 1; i < n; i++){
		A[i][i] = de[i];
		Redge(i){
			if ((ed[k].w >> p) & 1){
				A[i][n + 1] += 1.0;
				A[i][ed[k].to] += 1.0;
			}else A[i][ed[k].to] -= 1.0;
		}
	}
	A[n][n] = 1;
	gause();
	ans += (1 << p) * A[1][n + 1];
}
int main(){
	n = read(); m = read();
	int a,b,w;
	for (int i = 1; i <= m; i++){
		a = read(); b = read(); w = read();
		build(a,b,w);
	}
	for (p = 0; (1 << p) <= INF; p++) solve();
	printf("%.3lf\n",ans);
	return 0;
}