SRM 620 D2L3: RandomGraph, dp

 

It is also a question about probability. Considering the dp Method, the key is to find a breakthrough and convert the "at least one connected graph greater than 4" under the question condition into "All connected graphs are less than or equal to 3 ".

 

Code:

 

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                      using namespace std;#define CHECKTIME() printf(%.2lf, (double)clock() / CLOCKS_PER_SEC)typedef pair
                      
                        pii;typedef long long llong;typedef pair
                       
                         pll;#define mkp make_pair/*************** Program Begin **********************/double dp[51][51][51];bool solved[51][51][51];class RandomGraph {public:int n;double p;double rec(int a, int b, int c){double & res = dp[a][b][c];if (solved[a][b][c]) {return res;}int r = n - (a + 2 * b + 3 * c);// base caseif (0 == r) {res = 1.0;solved[a][b][c] = true;return res;}res = 0.0;// r != 0res += pow(1 - p, a + 2 * b + 3 * c) * rec(a + 1, b, c);if (a >= 1) {res += pow(1 - p, a + 2 * b + 3 * c - 1) * a * p * rec(a - 1, b + 1, c);}if (a >= 2) {res += pow(1 - p, a + 2 * b + 3 * c - 2) * p * p * a * (a - 1) / 2 * rec(a - 2, b, c + 1);}if (b >= 1) {res += pow(1 - p, a + 2 * b + 3 * c - 1) * p * 2 * b * rec(a, b - 1, c + 1);res += pow(1 - p, a + 2 * b + 3 * c - 2) * p * p * b * rec(a, b - 1, c + 1);}solved[a][b][c] = true;return res;}double probability(int n, int p) {double res = 0;this->n = n;this->p = p / 1000.0;memset(solved, 0, sizeof(solved));res = 1 - rec(0, 0, 0);return res;}};/************** Program End ************************/