bzoj1419: Red is good

题目链接

bzoj1419: Red is good

题解

由于有决策存在,所以要满足最有子结构，而期望期望正是能用来表示改状态的好坏
继续翻牌期望<0莫不如不取
设dp[i,j]表示有i张+1，j张-1的牌
dp[i,j]=max(0,(dp[i-1,j]+1)\frac{i}{i +j} + (dp[i,j-1]-1)\frac{j}{i+j})
dp[i,0]=i;
随便滚掉一维,空间就能过了

代码

#include<cstdio>
#include<algorithm> 
inline int read() { 
    int x = 0; 
    char c = getchar(); 
    while(c < '0' || c > '9')c = getchar(); 
    while(c <= '9' && c >= '0')x = x * 10 + c - '0',c = getchar(); 
    return x; 
} 
const int maxn = 5007; 
int R,B;  
double dp[2][maxn]; 
int main() { 
    R = read();B = read();  
    //for(int i = 1;i <= R;++ i) dp[i][0] = (double) i; 
    int k = 1;
    for(int i = 1;i <= R;++ i) { 
        k ^= 1;
        dp[k][0] = i;  
        for(int j = 1;j <= B;++ j) { 
            dp[k][j] = std::max((dp[k ^ 1][j] + 1.0) * ((double) i /(i + j)) + (dp[k][j - 1] - 1.0) * ((double) j / (i + j)),0.0);  
        }
    } 
    long long tmp = dp[k][B] * 1000000; 
    printf("%.6lf\n",(double) tmp / 1000000); 
    return 0; 
}