# BZOJ2337 [HNOI2011]XOR和路径

### Description

$n$个点的无向图，从$1$号点出发每次从所有相连的边随机选一个走过去，直到$n$号点结束。求路径上所有边权异或和的期望。$n\leq100, m\leq10000$，可能有重边自环。

### Code

#include <algorithm>
#include <cstdio>
const int N = 105;
const int M = 10005;
double A[N][N];
int n;
inline double abs(double x) { return x < 0 ? -x : x; }
double solve() {
for (int i = 1; i <= n; ++i) {
int j = i;
for (int k = i + 1; k <= n; ++k) if (abs(A[k][i]) > abs(A[j][i])) j = k;
for (int k = i; k <= n + 1; ++k) std::swap(A[i][k], A[j][k]);
for (int k = n + 1; k >= i; --k)
A[i][k] /= A[i][i];
for (int j = i + 1; j <= n; ++j)
for (int k = n + 1; k >= i; --k)
A[j][k] -= A[i][k] * A[j][i];
}
for (int i = n; i; --i)
for (int j = i - 1; j; --j)
A[j][n + 1] -= A[j][i] * A[i][n + 1];
return A[1][n + 1];
}
int u[M], v[M], w[M], deg[N];
int main() {
int m;
scanf("%d%d", &n, &m);
for (int i = 0; i < m; ++i) {
scanf("%d%d%d", &u[i], &v[i], &w[i]);
++deg[u[i]];
if (u[i] != v[i]) ++deg[v[i]];
}
double ans = .0;
for (int i = 0; i < 32; ++i) {
std::fill(A[0], A[n], .0);
for (int j = 1; j <= n; ++j) A[j][j] = 1.0;
for (int j = 0; j < m; ++j) {
if ((w[j] >> i) & 1) {
// f[u] += (1 - f[v]) / d[u]
A[u[j]][v[j]] += 1.0 / deg[u[j]];
A[u[j]][n + 1] += 1.0 / deg[u[j]];
if (u[j] != v[j]) {
A[v[j]][u[j]] += 1.0 / deg[v[j]];
A[v[j]][n + 1] += 1.0 / deg[v[j]];
}
} else {
// f[u] += f[v] / d[u]
A[u[j]][v[j]] -= 1.0 / deg[u[j]];
if (u[j] != v[j]) A[v[j]][u[j]] -= 1.0 / deg[v[j]];
}
}
std::fill(A[n], A[n + 1], .0);
A[n][n] = 1.0;
ans += solve() * (1 << i);
}
printf("%.3lf\n", ans);
return 0;
}

posted @ 2018-03-06 08:47  _rqy  阅读(91)  评论(0编辑  收藏  举报