CodeForces 148D 概率dp

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探讨了公主与龙通过概率DP算法决定新年夜活动的故事背景。在一个装有白鼠和黑鼠的袋子中，公主与龙轮流抽取，首先抽到白鼠者获胜。文章详细解析了这一过程的概率DP实现，包括状态转移方程的设计和实现细节。

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The dragon and the princess are arguing about what to do on the New Year’s Eve. The dragon suggests flying to the mountains to watch fairies dancing in the moonlight, while the princess thinks they should just go to bed early. They are desperate to come to an amicable agreement, so they decide to leave this up to chance.

They take turns drawing a mouse from a bag which initially contains w white and b black mice. The person who is the first to draw a white mouse wins. After each mouse drawn by the dragon the rest of mice in the bag panic, and one of them jumps out of the bag itself (the princess draws her mice carefully and doesn’t scare other mice). Princess draws first. What is the probability of the princess winning?

If there are no more mice in the bag and nobody has drawn a white mouse, the dragon wins. Mice which jump out of the bag themselves are not considered to be drawn (do not define the winner). Once a mouse has left the bag, it never returns to it. Every mouse is drawn from the bag with the same probability as every other one, and every mouse jumps out of the bag with the same probability as every other one.

Input
The only line of input data contains two integers w and b (0 ≤ w, b ≤ 1000).

Output
Output the probability of the princess winning. The answer is considered to be correct if its absolute or relative error does not exceed 10 - 9.

就是个简单的概率dp…
模拟着做做就过了…..
主要这种浮点数过不了要换编译器试试

#include<iostream>
#include<cmath>
#include<memory.h>
#include<algorithm>
#include<cstdio>
using namespace std;
double dp[1010][1010];
int main()
{
    int w,b;
    cin>>w>>b;
    if(w==0)
    {
        printf("%.10f\n",0);
        return 0;
    }
    for(int a=0;a<=w;a++)
    {
        for(int c=0;c<=b;c++)
        {
            if(c==0)
            {
                dp[a][c]=1;
                continue;
            }
            if(a==0)continue;
            dp[a][c]=1.0*a/(a+c)*1.0;
            if(c==2)
            {
                dp[a][c]+=1.0*(c*1.0/(a+c))*((c-1)*1.0/(a+c-1)*1.0)*dp[a-1][c-2];
                continue;
            }
            else if(c>=3)
            {
                dp[a][c]+=1.0*(c*1.0/(a+c)*1.0)*((c-1)*1.0/(a+c-1)*1.0)*((a*1.0*dp[a-1][c-2]*1.0+(c-2)*dp[a][c-3]*1.0)/(1.0*(a+c-2)*1.0));
                continue;
            }
        }
    }
    printf("%.10f\n",dp[w][b]);
    return 0;
}