[BZOJ 1419] Red is good

[题目链接]

[算法]

概率DP

用Fi,j表示还剩下i张红卡 ， j张黑卡 ， 期望获得的最大价值

时间复杂度 ： O(N ^ 2)

滚动数组 ， 将空间复杂度降至O(N)

[代码]

#include<bits/stdc++.h>
using namespace std;
#define MAXN 5010
#define RG register

int R , B;
double dp[2][MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
T f = 1; x = 0;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
x *= f;
}
int main()
{

dp[0][0] = 0;
for (RG int i = 0; i <= R; i++)
{
for (RG int j = 0; j <= B; j++)
{
if (i + j == 0) continue;
double p = 1.0 * i / (i + j);
if (i == 0) dp[i & 1][j] = max(0.0 , (dp[i & 1][j - 1] - 1) * (1 - p));
else if (j == 0) dp[i & 1][j] = max(0.0 , (dp[(i - 1) & 1][j] + 1) * p);
else dp[i & 1][j] = max(0.0 , 1.0 * (dp[(i - 1) & 1][j] + 1) *  p + 1.0 * (dp[i & 1][j - 1] - 1) * (1 - p));
}
}
dp[R & 1][B] *= (int)1e6;
dp[R & 1][B] = floor(dp[R & 1][B]);
dp[R & 1][B] /= (int)1e6;
printf("%.6lf\n" , dp[R & 1][B]);

return 0;

}

posted @ 2018-11-02 20:57  evenbao  阅读(140)  评论(0编辑  收藏  举报