Codeforces 148D(概率DP)

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Codeforces 148D
设$dp(i, j)$为轮到王妃选鼠的时候白鼠有$i$只，黑鼠有$j$只时赢的概率。
分4种情况.
1、王妃选到白，则概率为$\frac{i}{i+j}$
2、龙选到白，王妃输了，则概率为$0$
3、王妃选到黑，龙选到黑，逃了黑，转移到$dp(i, j-3)$，则概率为$\frac{j}{i+j} \times \frac{j-1}{i+j-1} \times \frac{j-2}{i+j-2}$
4、王妃选到黑，龙选到黑，逃了白，转移到$dp(i-1, j-2)$，则概率为$\frac{j}{i+j} \times \frac{j-1}{i+j-1} \times \frac{i}{i+j-2}$
那么综上所述可以得到状态转移方程
$dp(i,j) = \frac{i}{i+j} + \frac{j}{i+j} \times \frac{j-1}{i+j-1} \times \frac{j-2}{i+j-2} \times dp(i, j-3)+\frac{j}{i+j} \times \frac{j-1}{i+j-1} \times \frac{i}{i+j-2} \times dp(i-1, j-2)$
注意不要一起乘，不然溢出。

Codeforces Submission

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<queue>
#include<cmath>
#define ms(i, j) memset(i, j, sizeof i)
#define LL long long
#define db double
using namespace std;
const int MAXN = 1000 + 5;
int w, b;
db dp[MAXN][MAXN];
void clean() {
    ms(dp, 0);
}
void solve() {
    clean();
    for (int i=1;i<=w;i++) dp[i][0] = 1.0;
    for (int i=1;i<=w;i++) {
        for (int j=1;j<=b;j++) {
            dp[i][j] += (db)i / (db)(i + j);
            db p1, p2;
            if (j-3>=0) p1 = (db)j / (db)(i+j) * (db)(j-1) / (db)(i+j-1) * (db)(j-2) / (db)(i+j-2), dp[i][j] += dp[i][j-3] * p1;
            if (j-2>=0) p2 = (db)j / (db)(i+j) * (db)(j-1) / (db)(i+j-1) * (db)i / (db)(i+j-2), dp[i][j] += dp[i-1][j-2] * p2;
        }
    }
    printf("%.9f\n", dp[w][b]);
}
int main() {
    #ifndef ONLINE_JUDGE 
    freopen("1.in", "r", stdin);freopen("1.out", "w", stdout);
    #endif
    while(scanf("%d%d", &w, &b)==2) solve();
    return 0;
}

记忆化搜索写法：
Codeforces Submission

#include<cstdio>
#include<cstring>
#include<algorithm>
#define ms(i, j) memset(i, j, sizeof i)
#define LL long long
#define db double
#define FN2 "cf148d"
using namespace std;
const int MAXN = 1000 + 5;
db dp[MAXN][MAXN];
int w, b;
db dfs(int i, int j) {
    if (i <= 0) return 0;
    if (j <= 0) return 1;
    if (dp[i][j]>=0) return dp[i][j];
    dp[i][j] = 0;
    dp[i][j] += (db)i / (db)(i + j);
    if (j - 3 >= 0) {
        dp[i][j] += ((db)j / (db)(i + j) * (db)(j - 1) / (db)(i + j - 1) * (db)(j - 2) / (db)(i + j - 2)) * dfs(i, j - 3);
    }
    if (i - 1 >= 0  && j - 2 >= 0) {
        dp[i][j] += ((db)j / (db)(i + j) * (db)(j - 1) / (db)(i + j - 1) * (db)(i) / (db)(i + j - 2)) * dfs(i - 1, j - 2);
    }
    return dp[i][j];
}
void clean() {
    ms(dp, -1);
}
void solve() {
    clean();
    printf("%.9f\n", dfs(w, b));
}
int main() {
    #ifndef ONLINE_JUDGE
    freopen(FN2".in", "r", stdin); freopen(FN2".out", "w", stdout);
    #endif
    scanf("%d%d", &w, &b), solve();
    fclose(stdin); fclose(stdout);
    return 0;
}
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