题面
Sol
设\(f[i]\)表示炸弹到\(i\)不爆炸的期望
高斯消元即可
另外，题目中的概率\(p/q\)实际上为\(1-p/q\)
还有，谁能告诉我不加\(EPS\)，为什么会输出\(-0.00000\)
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(505);
const int __(100005);
IL ll Input(){
RG char c = getchar(); RG ll x = 0, z = 1;
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int n, m, fst[_], nxt[__], cnt, to[__], dg[_];
double ans, f[_], a[_][_], p, q;
IL void Add(RG int u, RG int v){
to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++; ++dg[v];
}
IL void Gauss(){
for(RG int i = 1; i < n; ++i)
for(RG int j = i + 1; j <= n; ++j){
RG double div = a[j][i] / a[i][i];
for(RG int k = 1; k <= n + 1; ++k) a[j][k] -= a[i][k] * div;
}
for(RG int i = n; i; --i){
f[i] = a[i][n + 1] / a[i][i];
for(RG int j = i - 1; j; --j) a[j][n + 1] -= f[i] * a[j][i];
}
}
int main(RG int argc, RG char* argv[]){
n = Input(); m = Input(); Fill(fst, -1);
q = Input(); q /= Input(); p = 1.0 - q;
for(RG int i = 1, u, v; i <= m; ++i)
u = Input(), v = Input(), Add(u, v), Add(v, u);
for(RG int u = 1; u <= n; u++){
a[u][u] = -1.0;
for(RG int e = fst[u]; e != -1; e = nxt[e]) a[to[e]][u] = p / dg[u];
}
a[1][n + 1] = -1.0; Gauss();
for(RG int i = 1; i <= n; ++i){
f[i] *= q;
if(fabs(f[i]) < 1e-9) f[i] = 0;
}
for(RG int i = 1; i <= n; ++i) printf("%.9lf\n", f[i]);
return 0;
}