HDU4089 Activation

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概率DP

状态转移方程：

dp[i][1] = p1*dp[i][1] + dp[i][i]*p2 + p4

dp[i][j] = p1*dp[i][j] + dp[i][j-1]*p2 + dp[i-1][j-1]*p3+p4 (j<=k)

dp[i][j] = p1*dp[i][j] + dp[i][j-1]*p2 + dp[i-1][j-1]*p3 (j>k)

一层一层的处理，然后对于每一层，依次带入转移方程或者解一个矩阵（或者是方程），把第一项求出来。第一项求出来之后后面的项根据状态转移方程依次求就行了。

#include <cstdio>#include <cstring>const double eps = 1e-6;const int maxn = 2005;int n, m, k;double p1, p2, p3, p4;double dp[maxn][maxn], p[maxn];int main(){    while (scanf("%d%d%d%lf%lf%lf%lf", &n, &m, &k, &p1, &p2, &p3, &p4) == 7)    {        if (p4 < eps) {puts("0.00000"); continue;}        p2 /= (1 - p1); p3 /= (1 - p1); p4 /= (1 - p1);        p[0] = 1;        for (int i=1;i<=n;i++) p[i] = p[i-1] * p2;        memset(dp, 0, sizeof(dp));        dp[1][1] = p4 / (p3 + p4);        for (int i=2;i<=n;i++)            for (int j=1;j<=n;j++)            {                if (j == 1)                {                    dp[i][1] = p4;                    for (int x=1;x<i;x++)                    {                        dp[i][1] += p[x] * p3 * dp[i-1][i-x];                        if (i+1-x <= k) dp[i][1] += p[x] * p4;                    }                    dp[i][1] /= (1 - p[i]);                }                else if (j <= k) dp[i][j] = dp[i][j-1] * p2 + dp[i-1][j-1] * p3 + p4;                else dp[i][j] = dp[i][j-1] * p2 + dp[i-1][j-1] * p3;            }        printf("%.5lf\n", dp[n][m]);    }    return 0;}