Hexo
0%

概率dp

Posted on In ,
Symbols count in article: 2.5k Reading time ≈ 2 mins.

概率dp

dp求概率

image-20240514220812741

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
int w, b;
double dp[1010][1010];

int main() {
  scanf("%d %d", &w, &b);
  memset(dp, 0, sizeof(dp));
  for (int i = 1; i <= w; i++) dp[i][0] = 1;  // 初始化
  for (int i = 1; i <= b; i++) dp[0][i] = 0;
  for (int i = 1; i <= w; i++) {
    for (int j = 1; j <= b; j++) {  // 以下为题面概率转移
      dp[i][j] += (double)i / (i + j);
      if (j >= 3) {
        dp[i][j] += (double)j / (i + j) * (j - 1) / (i + j - 1) * (j - 2) /
                    (i + j - 2) * dp[i][j - 3];
      }
      if (i >= 1 && j >= 2) {
        dp[i][j] += (double)j / (i + j) * (j - 1) / (i + j - 1) * i /
                    (i + j - 2) * dp[i - 1][j - 2];
      }
    }
  }
  printf("%.9lf\n", dp[w][b]);
  return 0;
}

dp求期望

image-20240514220306713

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
#include <cstdio>
using namespace std;
int n, s;
double dp[1010][1010];

int main() {
  scanf("%d %d", &n, &s);
  dp[n][s] = 0;
  for (int i = n; i >= 0; i--) {
    for (int j = s; j >= 0; j--) {
      if (i == n && s == j) continue;
      dp[i][j] = (dp[i][j + 1] * i * (s - j) + dp[i + 1][j] * (n - i) * j +
                  dp[i + 1][j + 1] * (n - i) * (s - j) + n * s) /
                 (n * s - i * j);  // 概率转移
    }
  }
  printf("%.4lf\n", dp[0][0]);
  return 0;
}

image-20240514220339671

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
#include <bits/stdc++.h>

using namespace std;

const int maxn = 2010;
int n, m, v, e;
int f[maxn][maxn], c[maxn], d[maxn];
double dp[maxn][maxn][2], p[maxn];

int main() {
  scanf("%d %d %d %d", &n, &m, &v, &e);
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &d[i]);
  for (int i = 1; i <= n; i++) scanf("%lf", &p[i]);
  for (int i = 1; i <= v; i++)
    for (int j = 1; j < i; j++) f[i][j] = f[j][i] = 1e9;

  int u, V, w;
  for (int i = 1; i <= e; i++) {
    scanf("%d %d %d", &u, &V, &w);
    f[u][V] = f[V][u] = min(w, f[u][V]);
  }

  for (int k = 1; k <= v; k++)
    for (int i = 1; i <= v; i++)  // 前面的，按照前面的题解进行一个状态转移
      for (int j = 1; j < i; j++)
        if (f[i][k] + f[k][j] < f[i][j]) f[i][j] = f[j][i] = f[i][k] + f[k][j];

  for (int i = 1; i <= n; i++)
    for (int j = 0; j <= m; j++) dp[i][j][0] = dp[i][j][1] = 1e9;

  dp[1][0][0] = dp[1][1][1] = 0;
  for (int i = 2; i <= n; i++)  // 有后效性方程
    for (int j = 0; j <= min(i, m); j++) {
      dp[i][j][0] = min(dp[i - 1][j][0] + f[c[i - 1]][c[i]],
                        dp[i - 1][j][1] + f[c[i - 1]][c[i]] * (1 - p[i - 1]) +
                            f[d[i - 1]][c[i]] * p[i - 1]);
      if (j != 0) {
        dp[i][j][1] = min(dp[i - 1][j - 1][0] + f[c[i - 1]][d[i]] * p[i] +
                              f[c[i - 1]][c[i]] * (1 - p[i]),
                          dp[i - 1][j - 1][1] +
                              f[c[i - 1]][c[i]] * (1 - p[i - 1]) * (1 - p[i]) +
                              f[c[i - 1]][d[i]] * (1 - p[i - 1]) * p[i] +
                              f[d[i - 1]][c[i]] * (1 - p[i]) * p[i - 1] +
                              f[d[i - 1]][d[i]] * p[i - 1] * p[i]);
      }
    }

  double ans = 1e9;
  for (int i = 0; i <= m; i++) ans = min(dp[n][i][0], min(dp[n][i][1], ans));
  printf("%.2lf", ans);

  return 0;
}