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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;

}

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