題目鏈接

BZOJ4418

題解

題意：從一個序列上某一點開始沿一個方向走，走到頭返回，每次走的步長各有概率，問走到一點的期望步數，或者無解

我們先將序列倍長形成循環序列，\(n = (N - 1) \times 2\)
按期望\(dp\)的套路，我們設\(f[i]\)為從\(i\)點出發到達終點的期望步數【一定要這麽做，不然轉移方程很難處理】，顯然終點\(f[Y] = f[(n - Y) \mod n] = 0\)
剩余的點
\[f[i] = \sum\limits_{j = 1}^{M} p_j(f[(i + j) \mod n] + j)\]
這是一個有後效性的轉移方程，高斯消元即可

但還沒完，有時候有些點是無法到達的，比如每次\(100 \%\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
 

#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
#define res register
#define eps 1e-9
using namespace std;
const int maxn = 205,maxm = 100005;
const double INF = 100000000000000000ll;
inline int read(){
    int out = 0,flag = 1; char
 
 c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
double A[maxn][maxn],p[maxn],ans[maxn];
int n,N,M,Y,X,D,vis[maxn];
int q[maxn],head,tail;
bool bfs(){
    for (res int i = 0; i < n; i++) vis[i] = false;
    vis[X] = true; q[head = tail = 0] = X;
    int u;
    while (head <= tail){
        u = q[head++];
        for (res int i = 1; i <= M; i++)
            if (p[i] > eps){
                int to = ((u + i) % n + n) % n;
                if (!vis[to]) vis[to] = true,q[++tail] = to;
            }
    }
    for (int i = 0; i < n; i++) if (!vis[i]){
        A[i][n] = INF,A[i][i] = 1;
    }
    return vis[Y] || vis[(n - Y) % n];
}
void pre(){
    for (int i = 0; i < n; i++)
        if (vis[i]){
            A[i][i] = 1;
            if (i == Y || i == (n - Y) % n) continue;
            for (int j = 1; j <= M; j++){
                int u = ((i +  j) % n + n) % n;
                A[i][u] += -p[j];
                A[i][n] += p[j] * j;
            }
        }
}
bool gause(){
    for (res int i = 0; i < n; i++){
        int j = i;
        for (res int k = i + 1; k < n; k++)
            if (fabs(A[k][i]) > fabs(A[j][i])) j = k;
        if (j != i) for (int k = i; k <= n; k++) swap(A[i][k],A[j][k]);
        if (fabs(A[i][i]) < eps) return false;
        for (res int j = i + 1; j < n; j++){
            double t = A[j][i] / A[i][i];
            for (res int k = i; k <= n; k++)
                A[j][k] -= A[i][k] * t;
        }
    }
    for (res int i = n - 1; ~i; i--){
        for (res int j = i + 1; j < n; j++)
            A[i][n] -= A[i][j] * ans[j];
        if (fabs(A[i][i]) < eps) return false;
        ans[i] = A[i][n] / A[i][i];
    }
    return true;
}
int main(){
    int T = read();
    while (T--){
        N = read(); M = read(); Y = read(); X = read(); D = read();
        n = 2 * (N - 1);
        if (D == -1){
            if (X == 0) D = 0;
            else D = 1;
        }
        if (D >= 1) X = (n - X) % n;
        for (int i = 1; i <= M; i++)
            p[i] = read() / 100.0;
        if (X == Y){puts("0.00"); continue;}
        for (res int i = 0; i < n; i++)
            for (res int j = 0; j <= n; j++)
                A[i][j] = 0;
        if (!bfs()) {puts("Impossible !"); continue;}
        pre();
        if (!gause() || ans[X] >= INF)
            puts("Impossible !");
        else printf("%.2lf\n",ans[X]);
    }
    return 0;
}

hdu4418 Time travel 【期望dp + 高斯消元】