标签:多少 abs line ifd out cto span getch vector
我们设\(f(i,j)\)是\((i,j)\)这个点期望被经过多少次
我们可以列出方程组来消元,由于终点只会被经过0次或者1次,期望就是概率
对于起点的话我们期望经过次数多加一个1
复杂度\(O(n^6)\)
#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <map>
//#define ivorysi
#define pb push_back
#define space putchar(‘ ‘)
#define enter putchar(‘\n‘)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define mo 974711
#define RG register
#define MAXN 200005
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {
res = 0;char c = getchar();T f = 1;
while(c < ‘0‘ || c > ‘9‘) {
if(c == ‘-‘) f = -1;
c = getchar();
}
while(c >= ‘0‘ && c <= ‘9‘) {
res = res * 10 + c - ‘0‘;
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) putchar(‘-‘),x = -x;
if(x >= 10) out(x / 10);
putchar(‘0‘ + x % 10);
}
int g[25][25],ind[25];
int N,M,a,b;
db P[25],f[505][505],ans[505];
int get_id(int x,int y) {
return (x - 1) * N + y;
}
void Guass() {
int T = N * N;
for(int i = 1 ; i <= T ; ++i) {
int l = i;
for(int j = i ; j <= T ; ++j) if(fabs(f[l][i]) < fabs(f[j][i])) l = j;
if(l != i) {
for(int j = 1 ; j <= T ; ++j) swap(f[l][j],f[i][j]);
}
for(int j = i + 1 ; j <= T ; ++j) {
db t = f[j][i] / f[i][i];
for(int k = i ; k <= T + 1; ++k) {
f[j][k] -= t * f[i][k];
}
}
}
for(int i = T ; i >= 1 ; --i) {
for(int j = i + 1 ; j <= T ; ++j) {
f[i][T + 1] -= f[i][j] * ans[j];
}
ans[i] = f[i][T + 1] / f[i][i];
}
}
void Solve() {
read(N);read(M);read(a);read(b);
int u,v;
for(int i = 1 ; i <= M ; ++i) {
read(u);read(v);
ind[u]++;ind[v]++;
g[u][v] = g[v][u] = 1;
}
for(int i = 1 ; i <= N ; ++i) scanf("%lf",&P[i]);
for(int i = 1 ; i <= N ; ++i) {
for(int j = 1 ; j <= N ; ++j) {
int t = get_id(i,j);
if(i == a && j == b) f[t][N * N + 1] += 1.0;
f[t][t] += 1.0;
for(int k = 1 ; k <= N ; ++k) {
if(k != i && !g[i][k]) continue;
for(int h = 1 ; h <= N ; ++h) {
if(j != h && !g[j][h]) continue;
if(k == h) continue;
db tmp = 1.0;
if(i == k) tmp *= P[i];
else tmp *= (1 - P[k]) / ind[k];
if(j == h) tmp *= P[h];
else tmp *= (1 - P[h]) / ind[h];
f[t][get_id(k,h)] += -tmp;
}
}
}
}
Guass();
for(int i = 1 ; i <= N ; ++i) {
printf("%.8lf ",ans[get_id(i,i)]);
}
enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
return 0;
}
标签:多少 abs line ifd out cto span getch vector
原文地址:https://www.cnblogs.com/ivorysi/p/9114725.html