HDU 4870 Rating 2014 Multi-University Training Contest 1 J题 概率DP+高斯消元
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题目大意:
就是现在有一个比赛,初始Rating是0, 每次参加比赛有P的概率Rating变为 Min(Rating + 50, 1000) 有(1 - P)的概率变为 Max(0, Rating -100)
为现在一个人用两个账号,每次选择用其中Rating较低的号参加比赛, 使得其中1个号的Rating达到1000所需要比赛的期望场数
大致思路:
很明显的概率DP, 由于变量数不多可以用高斯消元
不过这题的误差问题表示有些不能理解, Gauss消元误差较大
代码如下:
Result : Accepted Memory : 1812 KB Time : 1794 ms
/* * Author: Gatevin * Created Time: 2015/1/29 14:14:54 * File Name: Iris_Fleyja.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;double a[300][300];double x[300];int hash[22][22];int var, equ;/* * hash[i][j]为E[i][j]对应的变量编号 * 最终解的结果存在x[hash[i][j]]中 *//* * 如果用E[i][j] (i >= j)表示当前分数分别为i和j时到达目标状态的概率 * E[i][j] = P*(1 + E[i][j + 1]) + (1 - P)*(1 + E[i][Max(j - 2, 0)]) (i >= j) * 很明显E[i][j] = E[j][i], 所以高斯消元做的话只需要一半的变量就够了 * 另外感觉这题误差还是有些....不好说.. */int Gauss()//正常的高斯消元的想法{ for(int k = 0, col = 0; k < equ && col < var; k++, col++) { int max_r = k; for(int i = k + 1; i < equ; i++) if(fabs(a[i][col]) > fabs(a[max_r][col])) max_r = i; //if(fabs(a[max_r][col]) < eps) return 0; /* * 对于eps 为1e-8精度不够,方程组会被判无解 * 注释掉之后莫名过了... */ if(k != max_r) { for(int j = col; j < var; j++) swap(a[k][j], a[max_r][j]); swap(x[k], x[max_r]); } x[k] /= a[k][col]; for(int j = col + 1; j < var; j++) a[k][j] /= a[k][col]; a[k][col] = 1; for(int i = 0; i < equ; i++) if(i != k) { x[i] -= x[k]*a[i][col]; for(int j = col + 1; j < var; j++) a[i][j] -= a[k][j]*a[i][col]; a[i][col] = 0; } } return 1;}void buildEquation(double P)//建立方程组{ memset(a, 0, sizeof(a)); memset(x, 0, sizeof(x)); var = equ = 0; memset(hash, -1, sizeof(hash)); for(int i = 0; i <= 20; i++) { hash[20][i] = var++; x[equ] = 0; a[equ++][hash[20][i]] += 1; } for(int i = 0; i < 20; i++) for(int j = 0; j <= i; j++) { if(hash[i][j] == -1) hash[i][j] = var++; if(j + 1 <= i && hash[i][j + 1] == -1) hash[i][j + 1] = var++; if(j + 1 > i && hash[j + 1][i] == -1) hash[j + 1][i] = var++; if(j < 2) { if(hash[i][0] == -1) hash[i][0] = var++; a[equ][hash[i][0]] -= (1 - P); a[equ][hash[i][j]] += 1; if(j + 1 <= i) a[equ][hash[i][j + 1]] -= P; else a[equ][hash[j + 1][i]] -= P; x[equ++] = 1; } else { if(hash[i][j - 2] == -1) hash[i][j - 2] = var++; a[equ][hash[i][j - 2]] -= (1 - P); a[equ][hash[i][j]] += 1; if(j + 1 <= i) a[equ][hash[i][j + 1]] -= P; else a[equ][hash[j + 1][i]] -= P; x[equ++] = 1; } } return;}int main(){ double p; while(scanf("%lf", &p) != EOF) { buildEquation(p); Gauss(); printf("%.6f\n", x[hash[0][0]]); } return 0;}
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