hdu 5013 优化疑问+dp

http://acm.hdu.edu.cn/showproblem.php?pid=5013

m个游客，n座城市(m, n <= 16), 每个人从1走到n, 每次有一定概率停在原地，然后以后就不前进了。一个人到过一个城会得到一定的愉悦度，对于相邻的两座城，会额外产生Cj / Cj - 1 * Hj的愉悦度，Cj是到过j城的人数，Hj是到过j城的人在这里获得的愉悦度之和。求期望的总愉悦度。

根据题解给出的解法

http://blog.csdn.net/oilover/article/details/39526899

需要跑3s左右

优化成纯dp能变成78ms，但是看不懂别人的代码...

根据期望的线性性，分别求每个人每天的值的期望

设f[i][j][x][y]表示第i天，前j个人，前一天x，当前天y的概率

设g[i][j][x][y]表示第i天，前j个人，前一天x，当前天y的期望

那么

f[i][j][x][y]=f[i][j-1][x][y]*(1-p[j]^(i-2))+f[i][j-1][x-1][y]*p[j]^(i-2)*(1-p[j])+f[i][j-1][x-1][y-1]*p[j]^(i-1)

g[i][j][x][y]=g[i][j-1][x][y]*(1-p[j]^(i-2))+g[i][j-1][x-1][y]*p[j]^(i-2)*(1-p[j])+(g[i][j-1][x-1][y-1]+f[i][j-1][x-1][y-1]*h[j][i])*p[j]^(i-1)

状态方程不懂，求各位大神解释..

#include <cstdio>#include <cstdlib>#include <cmath>#include <cstring>#include <string>#include <queue>#include <stack>#include <iostream>#include <algorithm>using namespace std;#define RD(x) scanf("%d",&x)#define RD2(x,y) scanf("%d%d",&x,&y)#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define clr0(x) memset(x,0,sizeof(x))typedef long long LL;const int maxn = 20;int n, m;double p[maxn][maxn],h[maxn][maxn],f[maxn][maxn][maxn],g[maxn][maxn][maxn];int main() {    while (~RD2(m,n)) {        for (int i = 1; i <= m; i++) {            scanf("%lf", &p[i][1]);            p[i][0] = 1.0;            for (int j = 2; j <= n; j++) {                p[i][j] = p[i][j - 1] * p[i][1];            }        }        for (int i = 1; i <= m; i++) {            for (int j = 1; j <= n; j++) {                scanf("%lf", &h[i][j]);            }        }        clr0(f),clr0(g);        double ans = 0;        for (int i = 2; i <= n; i++) {            f[i][0][0] = 1;            for (int j = 1; j <= m; j++) {                for (int x = m; x >= 0; x--) {                    for (int y = x; y >= 0; y--) {                        f[i][x][y] *= (1.0 - p[j][i - 2]);                        if (x > 0) {                            f[i][x][y] += f[i][x - 1][y] * p[j][i - 2] * (1.0 - p[j][1]);                            if (y > 0) {                                f[i][x][y] += f[i][x - 1][y - 1] * p[j][i - 1];                            }                        }                        g[i][x][y] *= (1.0 - p[j][i - 2]);                        if (x > 0) {                            g[i][x][y] += g[i][x - 1][y] * p[j][i - 2] * (1.0 - p[j][1]);                            if (y > 0) {                                g[i][x][y] += (g[i][x - 1][y - 1] + f[i][x - 1][y - 1] * h[j][i]) * p[j][i - 1];                            }                        }                    }                }            }            for (int x = 1; x <= m; x++) {                for (int y = 1; y <= x; y++) {                    ans += g[i][x][y] * ((double)y / x + 1.0);                }            }        }        for (int i = 1; i <= m; i++) {            ans += h[i][1];        }        printf("%.10f\n", ans);    }    return 0;}