bzoj 3036 绿豆蛙的归宿 期望dp
题面
解法
\[f_x=\sum \frac{f_y+w(x,y)}{out_x}$
因为是一个DAG，直接记忆化即可
时间复杂度：$O(n+m)$
### 代码
```cpp
#include <bits/stdc++.h>
#define N 100010
using namespace std;
template <typename node> void chkmax(node &x, node y) {x = max(x, y);}
template <typename node> void chkmin(node &x, node y) {x = min(x, y);}
template <typename node> void read(node &x) {
x = 0; int f = 1; char c = getchar();
while (!isdigit(c)) {if (c == '-') f = -1; c = getchar();}
while (isdigit(c)) x = x * 10 + c - '0', c = getchar(); x *= f;
}
struct Edge {
int next, num, v;
} e[N * 5];
int n, m, cnt, s[N];
double f[N];
void add(int x, int y, int v) {
e[++cnt] = (Edge) {e[x].next, y, v};
e[x].next = cnt;
}
double dp(int x) {
if (x == n) return 0;
if (f[x]) return f[x];
for (int p = e[x].next; p; p = e[p].next) {
int k = e[p].num, v = e[p].v;
f[x] += (dp(k) + v) / s[x];
}
return f[x];
}
int main() {
read(n), read(m); cnt = n;
for (int i = 1; i <= m; i++) {
int x, y, v;
read(x), read(y), read(v);
add(x, y, v); s[x]++;
}
cout << fixed << setprecision(2) << dp(1) << "\n";
return 0;
}
```\]
