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;
}

hdu 5013 优化疑问+dp