[USACO10HOL]赶小猪

嘟嘟嘟


这题和某一类概率题一样，大体思路都是高斯消元解方程。


不过关键还是状态得想明白。刚开始令\(f[i]\)表示炸弹在点\(i\)爆的概率，然后发现这东西根本无法转移（或者说概率本来就是\(\frac{p}{q}\)？），于是就考虑换状态。
一个非常好的状态是炸弹传到点\(i\)的概率，这样答案再乘以一个\(\frac{p}{q}\)就好了。转移也特别好办：\(f[i] = \sum (1 - \frac{p}{q}) * \frac{1}{du[v]}*f[v]\)。
别忘了结点1的概率要加1。
于是这题就做完了。

#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 = 305;
const int maxe = 1e6 + 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, m, p, q;
db P;
struct Edge
{
  int nxt, to;
}e[maxe];
int head[maxn], ecnt = -1, du[maxn];
In void addEdge(int x, int y)
{
  ++du[x];
  e[++ecnt] = (Edge){head[x], y};
  head[x] = ecnt;
}

db f[maxn][maxn], ans[maxn];
In void init()
{
  f[1][n + 1] = 1;
  for(int i = 1; i <= n; ++i)
    {
      f[i][i] = 1;
      for(int j = head[i], v; ~j; j = e[j].nxt)
    f[v = e[j].to][i] -= (1 - P) * 1.0 / du[i];
    }
}
In void Gauss()
{
  for(int i = 1; i <= n; ++i)
    {
      int pos = i;
      for(int j = i + 1; j <= n; ++j)
    if(fabs(f[j][i]) > fabs(f[pos][i])) pos = j;
      if(pos ^ i) swap(f[pos], f[i]);
      if(fabs(f[i][i]) < eps) continue;
      db tp = f[i][i];
      for(int j = i; j <= n + 1; ++j) f[i][j] /= tp;
      for(int j = i + 1; j <= n; ++j)
    {
      db tp = f[j][i];
      for(int k = i; k <= n + 1; ++k) f[j][k] -= tp * f[i][k];
    }
    }
  for(int i = n; i; --i)
    {
      ans[i] = f[i][n + 1];
      for(int j = i - 1; j; --j) f[j][n + 1] -= f[j][i] * ans[i];
    }
}

int main()
{
  //MYFILE();
  Mem(head, -1);
  n = read(), m = read(), p = read(), q = read();
  P = 1.0 * p / q;
  for(int i = 1; i <= m; ++i)
    {
      int x = read(), y = read();
      addEdge(x, y), addEdge(y, x);
    }
  init(), Gauss();
  for(int i = 1; i <= n; ++i) printf("%.9lf\n", ans[i] * P);
  return 0;
}