[hdu4870]高斯消元

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。


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/* ******************************************************************************** */
#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;
}
/* ******************************************************************************** */

 

posted @ 2015-08-03 04:26  jklongint  阅读(205)  评论(0)    收藏  举报