UvaLive6441（期望概率dp）

Posted 2021-02-14 alphawa

1.涉及负数时同时维护最大和最小，互相转移。

2.考场上最大最小混搭转移WA，赛后发现如果是小的搭小的，大的搭大的就可过，类似这种：

db a = (C[i] - W[i]) * dp1[i - 1][j - 1] - (1 - C[i] - W[i]) * dp1[i - 1][j - 1];
db b = (C[i] - W[i]) * dp2[i - 1][j - 1] - (1 - C[i] - W[i]) * dp2[i - 1][j - 1];

于是猜测原因的话可能是：最大和最小的不一定是同一种方案，而P1、P2、P3这三种情况转移到现态必须是同一种方案转移过来？

 

 1 const int maxn = 2e3 + 5;
 2 int n, m, T;
 3 db dp1[maxn][maxn], dp2[maxn][maxn], C[maxn], W[maxn];
 4 
 5 int main() {
 6     scanf("%d", &T);
 7     for (int kase = 1; kase <= T; kase++) {
 8         scanf("%d %d", &n, &m);
 9         rep(i, 1, n) {
10             scanf("%lf", &C[i]);
11             C[i] /= 100;
12         }
13         rep(i, 1, n) {
14             scanf("%lf", &W[i]);
15             W[i] /= 100;
16         }
17         init(dp1, 0);
18         init(dp2, 0);
19         dp1[0][0] = dp2[0][0] = 15000;
20         for (int i = 1; i <= n; i++) {
21             for (int j = 0; j <= m; j++) {
22                 if (j > i)    break;
23                 dp1[i][j] = 15001, dp2[i][j] = -15001;
24                 if (j < i) {
25                     db a = (1 - 2.0*W[i]) * dp1[i - 1][j];
26                     db b = (1 - 2.0*W[i]) * dp2[i - 1][j];
27                     dp1[i][j] = min(a, b);
28                     dp2[i][j] = max(a, b);
29                 }
30                 if (j) {
31                     db a = (C[i] - W[i]) * dp1[i - 1][j - 1] - (1 - C[i] - W[i]) * dp1[i - 1][j - 1];
32                     db b = (C[i] - W[i]) * dp2[i - 1][j - 1] - (1 - C[i] - W[i]) * dp2[i - 1][j - 1];
33                     dp1[i][j] = min(dp1[i][j], min(a, b));
34                     dp2[i][j] = max(dp2[i][j], max(a, b));
35                 }
36             }
37         }
38         db ans = 15001;
39         for (int j = 0; j <= m; j++)    ans = min(ans, dp1[n][j]);
40         printf("Case #%d: %.3lf
", kase, ans);
41     }
42     return 0;
43 }