[SHOI2014]概率充电器

嘟嘟嘟


这种复杂的概率大题我果然是每做出来……


然后我找到了一篇极棒的题解，小学生都能看懂（大佬就是大佬啊）：题解 P4284 【[SHOI2014]概率充电器】，第二次dp的状态方程真的很妙啊。


刚开始我总按照套路想设(dp[u])表示(u)的子树的期望，看完题解后发现这是没有依据的，因为每一个元件可以直接充电，不一定要形成依赖关系。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
#include<assert.h>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 5e5 + 5;
In ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - '0', ch = getchar();
  if(last == '-') ans = -ans;
  return ans;
}
In void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
In void MYFILE()
{
#ifndef mrclr
  freopen(".in", "r", stdin);
  freopen(".out", "w", stdout);
#endif
}

int n, a[maxn];
struct Edge
{
  int nxt, to; db p;
}e[maxn << 1];
int head[maxn], ecnt = -1;
In void addEdge(int x, int y, int w)
{
  e[++ecnt] = (Edge){head[x], y, 0.01 * w};
  head[x] = ecnt;
}

db dp[maxn];
In void dfs1(int now, int _f)
{
  dp[now] = 0.01 * a[now];
  for(int i = head[now], v; ~i; i = e[i].nxt)
    {
      if((v = e[i].to) == _f) continue;
      dfs1(v, now);
      dp[now] = dp[now] + dp[v] * e[i].p - dp[now] * dp[v] * e[i].p;
    }
}
db ans = 0;
In void dfs2(int now, int _f)
{
  ans += dp[now];
  for(int i = head[now], v; ~i; i = e[i].nxt)
    {
      if((v = e[i].to) == _f) continue;
      db x = dp[v] * e[i].p;
      if(x < 1)
	{
	  db tp = (dp[now] - x) / (1.0 - x);
	  dp[v] = dp[v] + tp * e[i].p - dp[v] * tp * e[i].p;
	}
      dfs2(v, now);
    }
}

int main()
{
  //MYFILE();
  Mem(head, -1);
  n = read();
  for(int i = 1; i < n; ++i)
    {
      int x = read(), y = read(), p = read();
      addEdge(x, y, p), addEdge(y, x, p);
    }
  for(int i = 1; i <= n; ++i) a[i] = read();
  dfs1(1, 0), dfs2(1, 0);
  printf("%.6lf
", ans);
  return 0;
}