HDU 4089.Activation

题目:http://acm.hdu.edu.cn/showproblem.php?pid=4089

AC代码(C++):

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

#define INF 0x3f3f3f3f
#define eps 1e-5
typedef unsigned long long ull;
typedef long long ll;
typedef __int64 I64d;

using namespace std;

int n, m, k;
double p1, p2, p3, p4;
double dp[2005][2005];

int main() {
    while (cin >> n >> m >> k >> p1 >> p2 >> p3 >> p4) {
        if (p4 < eps) {
            cout << "0.00000" << endl;
            continue;
        }
        double p01 = p2 / (1 - p1);
        double p02 = p3 / (1 - p1);
        double p03 = p4 / (1 - p1);
        dp[1][1] = p03 / (1 - p01);
        for (int i = 2; i <= n; i++) {
            dp[i][i] = 0;
            dp[i][i] += pow(p01, i - 1)*p03;
            for (int j = 2; j <= k&&j <= i; j++)dp[i][i] += pow(p01, i - j) * (p02*dp[i - 1][j - 1] + p03);
            for (int j = k + 1; j <= i; j++)dp[i][i] += pow(p01, i - j)*p02*dp[i - 1][j - 1];
            dp[i][i] /= (1 - pow(p01, i));
            dp[i][1] = p01*dp[i][i] + p03;
            for (int j = 2; j <= k&&j < i; j++)dp[i][j] = p01*dp[i][j - 1] + p02*dp[i - 1][j - 1] + p03;
            for (int j = k + 1; j < i; j++)dp[i][j] = p01*dp[i][j - 1] + p02*dp[i - 1][j - 1];
        }
        printf("%.5lf\n", dp[n][m]);
    }

    //system("pause");
}

总结: 动态规划, 这类带环的表达式要通过化简和迭代(即先求出dp[i][i])来做. 这题有个坑, 算了多余的j不但会答案错误, 还会超内存.