【BZOJ1778】[Usaco2010 Hol]Dotp 驱逐猪猡【概率DP】【高斯消元】

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【题目链接】

竟然被卡eps。

设xu为到达u点没有爆炸的概率。

方程为xu - (1 - p / q)∑xv / du[v] = [u = 1]。

高斯消元一发就好了。

答案要乘p / q。

/* Footprints In The Blood Soaked Snow */#include <cstdio>#include <cmath>#include <algorithm>using namespace std;typedef long double LD;const int maxn = 305, maxm = 50005;const LD eps = 1e-13;int n, m, head[maxn], cnt, du[maxn];LD p, q, A[maxn][maxn];struct _edge {int v, next;} g[maxm << 1];inline int iread() {int f = 1, x = 0; char ch = getchar();for(; ch < '0' || ch > '9'; ch = getchar()) f = ch == '-' ? -1 : 1;for(; ch >= '0' && ch <= '9'; ch = getchar()) x = x * 10 + ch - '0';return f * x;}inline void add(int u, int v) {g[cnt] = (_edge){v, head[u]};head[u] = cnt++;}inline void gauss() {for(int i = 1; i <= n; i++) {if(fabs(A[i][i]) < eps) for(int j = i + 1; j <= n; j++) if(fabs(A[j][i]) > eps) {for(int k = 1; k <= m; k++) swap(A[i][k], A[j][k]);break;}for(int j = 1; j <= n; j++) if(i ^ j && fabs(A[j][i]) > eps) {LD t = A[j][i] / A[i][i];for(int k = 1; k <= m; k++) A[j][k] -= t * A[i][k];}}for(int i = 1; i <= n; i++) A[i][m] /= A[i][i];}int main() {n = iread(); m = iread(); p = iread(); q = iread(); LD pro = p / q, antipro = 1.0 - pro;for(int i = 1; i <= n; i++) head[i] = -1; cnt = 0;for(int i = 1; i <= m; i++) {int u = iread(), v = iread();add(u, v); add(v, u);du[u]++; du[v]++;}m = n + 1;for(int i = 1; i <= n; i++) {A[i][i] = 1.0;for(int j = head[i]; ~j; j = g[j].next) A[i][g[j].v] = -antipro / (LD)du[g[j].v];}A[1][m] = 1.0;gauss();for(int i = 1; i <= n; i++) printf("%.9Lf\n", A[i][m] * pro);return 0;}

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