[USACO10HOL]赶小猪Driving Out the Piggies 题解
题目地址:洛谷:【P2973】[USACO10HOL]赶小猪Driving Out the Piggi… – 洛谷、BZOJ:Problem 1778. — [Usaco2010 Hol]Dotp 驱逐猪猡
题目描述
一个无向图,节点1有一个炸弹,在每个单位时间内,有p/q的概率在这个节点炸掉,有1-p/q的概率随机选择一条出去的路到其他的节点上。问最终炸弹在每个节点上爆炸的概率。
输入输出格式
输入格式:
* Line 1: Four space separated integers: N, M, P, and Q
* Lines 2..M+1: Line i+1 describes a road with two space separated integers: A_j and B_j
输出格式:
* Lines 1..N: On line i, print the probability that city i will be destroyed as a floating point number. An answer with an absolute error of at most 10^-6 will be accepted (note that you should output at least 6 decimal places for this to take effect).
输入输出样例
输入样例#1:
2 1 1 2 1 2
输出样例#1:
0.666666667 0.333333333
题解
我们考虑每次到达每个点爆炸是等可能的,因此在每个点爆炸的概率之比等于到达每个点的期望次数之比。而到达每个点的期望次数可以通过以下式子算出
\mathrm{E}(u) = \sum_{(u, v) \in E} \frac{\mathrm{E}(v) \times (1 - p / q)}{\mathrm{degree}[v]}
由于原图不一定是树,无法直接转移,我们考虑高斯消元的解法。对于除了1的节点,到达该点的期望次数等于右边这个求和,而对于节点1,由于一开始就在1位置,次数多1,方程等号右边就是1。
我们求出的是期望次数,和不为1,而概率和为1,需要加起来除一下才能得到答案。
代码
// Code by KSkun, 2018/3
#include <cstdio>
#include <cmath>
#include <algorithm>
typedef long long LL;
inline char fgc() {
static char buf[100000], *p1 = buf, *p2 = buf;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++;
}
inline LL readint() {
register LL res = 0, neg = 1;
char c = fgc();
while(c < '0' || c > '9') {
if(c == '-') neg = -1;
c = fgc();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = fgc();
}
return res * neg;
}
const int MAXN = 305, EPS = 1e-10;
int n, m, u, v, p, q, gra[MAXN][MAXN], deg[MAXN];
double mat[MAXN][MAXN];
inline void gauss() {
for(int i = 1; i <= n; i++) {
int r = i;
for(int j = i + 1; j <= n; j++) {
if(std::fabs(mat[r][i]) < std::fabs(mat[j][i])) r = j;
}
if(r != i) {
for(int j = 1; j <= n + 1; j++) {
std::swap(mat[r][j], mat[i][j]);
}
}
for(int j = 1; j <= n; j++) {
if(j != i) {
double t = mat[j][i] / mat[i][i];
for(int k = i + 1; k <= n + 1; k++) {
mat[j][k] -= mat[i][k] * t;
}
}
}
}
for(int i = 1; i <= n; i++) {
mat[i][n + 1] /= mat[i][i];
}
}
int main() {
n = readint(); m = readint(); p = readint(); q = readint();
while(m--) {
u = readint(); v = readint();
gra[u][v] = gra[v][u] = 1;
deg[u]++; deg[v]++;
}
for(int i = 1; i <= n; i++) {
mat[i][i] = 1;
for(int j = 1; j <= n; j++) {
if(gra[i][j]) mat[j][i] = -(1 - double(p) / q) / deg[i];
}
}
mat[1][n + 1] = 1;
gauss();
double sum = 0;
for(int i = 1; i <= n; i++) {
sum += mat[i][n + 1];
}
for(int i = 1; i <= n; i++) {
printf("%.9lf\n", mat[i][n + 1] / sum);
}
return 0;
}