You can not select more than 25 topics
Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
|
|
|
|
#include <bits/stdc++.h>
|
|
|
|
|
using namespace std;
|
|
|
|
|
const int N = 1e5 + 10, M = 2 * N;
|
|
|
|
|
|
|
|
|
|
//邻接表
|
|
|
|
|
int h[N], e[M], ne[M], w[M], idx;
|
|
|
|
|
void add(int a, int b, int c) {
|
|
|
|
|
e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
int n, m; // n个顶点,m条边
|
|
|
|
|
|
|
|
|
|
int out[N], in[N]; //出度,入度
|
|
|
|
|
double f[N], g[N]; // f:数学期望结果 g:概率
|
|
|
|
|
|
|
|
|
|
void topsort() {
|
|
|
|
|
queue<int> q;
|
|
|
|
|
//起点为1,起点的概率为100%
|
|
|
|
|
q.push(1);
|
|
|
|
|
g[1] = 1.0;
|
|
|
|
|
f[1] = 0.0;
|
|
|
|
|
|
|
|
|
|
// DAG,执行拓扑序,以保证计算的顺序正确,确保递归过程中,前序数据都已处理完毕
|
|
|
|
|
while (q.size()) {
|
|
|
|
|
auto u = q.front();
|
|
|
|
|
q.pop();
|
|
|
|
|
|
|
|
|
|
for (int i = h[u]; ~i; i = ne[i]) { //枚举的是每边相邻边
|
|
|
|
|
int j = e[i]; //此边,一端是t,另一端是j
|
|
|
|
|
//此边边条w[i]
|
|
|
|
|
f[j] += (f[u] + w[i] * g[u]) / out[u];
|
|
|
|
|
g[j] += g[u] / out[u]; // p[j]也需要概率累加
|
|
|
|
|
//拓扑序的标准套路
|
|
|
|
|
in[j]--;
|
|
|
|
|
if (!in[j]) q.push(j);
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
}
|
|
|
|
|
|
|
|
|
|
int main() {
|
|
|
|
|
//初始化邻接表
|
|
|
|
|
memset(h, -1, sizeof h);
|
|
|
|
|
cin >> n >> m;
|
|
|
|
|
|
|
|
|
|
while (m--) {
|
|
|
|
|
int a, b, c;
|
|
|
|
|
cin >> a >> b >> c;
|
|
|
|
|
add(a, b, c);
|
|
|
|
|
//维护出度,入度
|
|
|
|
|
out[a]++, in[b]++;
|
|
|
|
|
}
|
|
|
|
|
//拓扑序
|
|
|
|
|
topsort();
|
|
|
|
|
|
|
|
|
|
//正向递推,输出结果,保留两位小数
|
|
|
|
|
printf("%.2lf", f[n]);
|
|
|
|
|
|
|
|
|
|
return 0;
|
|
|
|
|
}
|