SRM 620 D2L3: RandomGraph, dp

题目：http://community.topcoder.com/stat?c=problem_statement&pm=13143&rd=15853

参考：http://apps.topcoder.com/wiki/display/tc/SRM+620

又是一道关于概率的题目，考虑dp方法，关键是找准突破口，将题目条件的“至少有一个大于4的连通图”转换为“所有连通图都小于等于3”。

代码：

#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <iostream>
#include <sstream>
#include <iomanip>

#include <bitset>
#include <string>
#include <vector>
#include <stack>
#include <deque>
#include <queue>
#include <set>
#include <map>

#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cmath>
#include <cstring>
#include <ctime>
#include <climits>
using namespace std;

#define CHECKTIME() printf("%.2lf\n", (double)clock() / CLOCKS_PER_SEC)
typedef pair<int, int> pii;
typedef long long llong;
typedef pair<llong, llong> 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 case
if (0 == r) {
res = 1.0;
solved[a][b][c] = true;
return res;
}
res = 0.0;
// r != 0
res += 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 ************************/