UvaLive6441（期望概率dp）

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

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？

 const int maxn = 2e3 + ;
 int n, m, T;
 db dp1[maxn][maxn], dp2[maxn][maxn], C[maxn], W[maxn];

 int main() {
     scanf("%d", &T);
     for (int kase = ; kase <= T; kase++) {
         scanf("%d %d", &n, &m);
         rep(i, , n) {
             scanf("%lf", &C[i]);
             C[i] /= ;
         }
         rep(i, , n) {
             scanf("%lf", &W[i]);
             W[i] /= ;
         }
         init(dp1, );
         init(dp2, );
         dp1[][] = dp2[][] = ;
         for (int i = ; i <= n; i++) {
             for (int j = ; j <= m; j++) {
                 if (j > i)    break;
                 dp1[i][j] = , dp2[i][j] = -;
                 if (j < i) {
                     db a = ( - 2.0*W[i]) * dp1[i - ][j];
                     db b = ( - 2.0*W[i]) * dp2[i - ][j];
                     dp1[i][j] = min(a, b);
                     dp2[i][j] = max(a, b);
                 }
                 if (j) {
                     db a = (C[i] - W[i]) * dp1[i - ][j - ] - ( - C[i] - W[i]) * dp1[i - ][j - ];
                     db b = (C[i] - W[i]) * dp2[i - ][j - ] - ( - C[i] - W[i]) * dp2[i - ][j - ];
                     dp1[i][j] = min(dp1[i][j], min(a, b));
                     dp2[i][j] = max(dp2[i][j], max(a, b));
                 }
             }
         }
         db ans = ;
         for (int j = ; j <= m; j++)    ans = min(ans, dp1[n][j]);
         printf("Case #%d: %.3lf\n", kase, ans);
     }
     return ;
 }