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题意：小明有2个账号，rating都是0分，每打一场赢的概率为P，假设当前分为x，赢了分数变为min(1000,x+50)，输了则分数变为max(0,x-100)，小明每次都选rating小的账号打，求打到有一个账号为1000所需的场数的期望值
思路：很明显需要把分数离散化，50分为1个单位。利用期望的可加性建立状态：dp(x,y)(x<=y))表示当前两个账号rating小的为x，大的为y，到达目标状态所需场数的期望值，则有dp(x,y)=P*dp(x+1,y)+（1-P)*dp(x-1,y){这里为了描述方便，没有考虑边界}，dp(19,20)=0。建好图后由于所有的dp值都是未知的，但转移时的每一项前面的系数为常数，所以可以考虑用高斯消元来确定每个状态的值。
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 | /* ******************************************************************************** */#include <iostream> //#include <cstdio> //#include <cmath> //#include <cstdlib> //#include <cstring> //#include <vector> //#include <ctime> //#include <deque> //#include <queue> //#include <algorithm> //#include <map> //#include <cmath> //using namespace std; // //#define pb push_back //#define mp make_pair //#define X first //#define Y second //#define all(a) (a).begin(), (a).end() //#define fillchar(a, x) memset(a, x, sizeof(a)) // //typedef pair<int, int> pii; //typedef long long ll; //typedef unsigned long long ull; // //#ifndef ONLINE_JUDGE //void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} //void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> //void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; //while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> //void print(const T t){cout<<t<<endl;}template<typename F,typename...R> //void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> //void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} //#endif // ONLINE_JUDGE //template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} //template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} //template<typename T> //void V2A(T a[],const vector<T>&b){for(int i=0;i<b.size();i++)a[i]=b[i];} //template<typename T> //void A2V(vector<T>&a,const T b[]){for(int i=0;i<a.size();i++)a[i]=b[i];} // //const double PI = acos(-1.0); //const int INF = 1e9 + 7; // ///* -------------------------------------------------------------------------------- */struct Gauss { const static int maxn = 1e3 + 7; double A[maxn][maxn]; int n; double* operator [] (int x) { return A[x]; } /** 要求系数矩阵可逆 A是增广矩阵，A[i][n]是第i个方程右边的常数bi 运行结束后 A[i][n]是第i个未知数的值 **/ void solve() { for (int i = 0; i < n; i ++) { int r = i; for (int j = i + 1; j < n; j ++) { if (fabs(A[j][i]) > fabs(A[r][i])) r = j; } if (r != i) for (int j = 0; j <= n; j ++) swap(A[r][j], A[i][j]); for (int j = n; j >= i; j --) { for (int k = i + 1; k < n; k ++) { A[k][j] -= A[k][i] / A[i][i] * A[i][j]; } } } for(int i = n - 1; i >= 0; i --) { for (int j = i + 1; j < n; j ++) { A[i][n] -= A[j][n] * A[i][j]; } A[i][n] /= A[i][i]; } }};Gauss solver;int c, n;double p;int hsh[22][22];void add(int x, int y, int z) { solver[c][x] += 1; solver[c][y] += -p; solver[c ++][z] += p - 1;}void init() { fillchar(solver.A, 0); c = 0; for (int i = 0; i < 20; i ++) { for (int j = i; j < 20; j ++) { int win = i + 1, unwin = max(i - 2, 0); add(hsh[i][j], hsh[min(win, j)][max(win, j)], hsh[unwin][j]); } } solver[c ++][hsh[19][20]] = 1; for (int i = 0; i < c - 1; i ++) { solver[i][n] = 1; }}void pre_init() { int c = 0; for (int i = 0; i < 20; i ++) { for (int j = i; j < 20; j ++) { hsh[i][j] = c ++; } } hsh[19][20] = c ++; solver.n = n = c;}int main() {#ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout);#endif // ONLINE_JUDGE pre_init(); while (cin >> p) { init(); solver.solve(); printf("%.10f\n", solver[0][n]); } return 0;}/* ******************************************************************************** */ |
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原文地址：http://www.cnblogs.com/jklongint/p/4697237.html
