cf113D. Museum(期望 高斯消元)

题意

题目链接

Sol

设\(f[i][j]\)表示Petya在\(i\)，\(Vasya\)在\(j\)的概率，我们要求的是\(f[i][i]\)

直接列方程高斯消元即可，由于每个状态有两维，因此时间复杂度为\(O(n^6)\)

注意不能从终止节点转移而来

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 2333;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M, a, b, Lim;
double p[MAXN], f[MAXN][MAXN], E[MAXN][MAXN], deg[MAXN];
vector<int> v[MAXN];
int id[MAXN][MAXN], tot;
void Pre() {
    f[id[a][b]][Lim + 1] = -1;
    for(int i = 1; i <= N; i++) {
        for(int j = 1; j <= N; j++) {
            int now = id[i][j];
            --f[now][now];//tag
            if(i != j) f[now][now] += p[i] * p[j];
            for(auto &x : v[i]) {
                for(auto &y : v[j]) {
                    if(x == y) continue;
                    int nxt = id[x][y];
                    f[now][nxt] += (1.0 - p[x]) * (1.0 - p[y]) / deg[x] / deg[y];
                }
            }
            for(auto &x : v[i]) {
                int nxt = id[x][j];
                if(x == j) continue;
                f[now][nxt] += (1.0 - p[x]) * p[j] / deg[x];
            }
            for(auto &y : v[j]) {
                int nxt = id[i][y];
                if(i == y) continue;
                f[now][nxt] += p[i] * (1.0 - p[y]) / deg[y];
            }
        }
    }
}
void Gauss() {
    for(int i = 1; i <= Lim; i++) {
        int mx = i;
        for(int j = i + 1; j <= Lim; j++) if(f[j][i] > f[mx][i] && f[j][i] != 0) swap(j, mx);
        if(mx != i) swap(f[i], f[mx]);
    //  assert(fabs(f[i][i] < 1e-13));
        for(int j = 1; j <= Lim; j++) {
            if(i == j) continue;
            double p = f[j][i] / f[i][i];
            for(int k = i; k <= Lim + 1; k++) f[j][k] -= f[i][k] * p;
        }
    }
    for(int i = 1; i <= Lim; i++) f[i][i] = f[i][Lim + 1] / f[i][i];
}
int main() {
    N = read(); M = read(); a = read(); b = read(); Lim = N * N;
    for(int i = 1; i <= M; i++) {
        int x = read(), y = read();
        v[x].push_back(y); v[y].push_back(x);
        deg[x]++; deg[y]++;
    }
    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++)
            id[i][j] =  ++tot;
    Pre();
    Gauss();
    for(int i = 1; i <= N; i++) printf("%.10lf ", f[id[i][i]][id[i][i]]);
    return 0;
}