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 - 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？

const int maxn = 2e3 + 5;
int n, m, T;
db dp1[maxn][maxn], dp2[maxn][maxn], C[maxn], W[maxn];

int main() {
    scanf("%d", &T);
    for (int kase = 1; kase <= T; kase++) {
        scanf("%d %d", &n, &m);
        rep(i, 1, n) {
            scanf("%lf", &C[i]);
            C[i] /= 100;
        }
        rep(i, 1, n) {
            scanf("%lf", &W[i]);
            W[i] /= 100;
        }
        init(dp1, 0);
        init(dp2, 0);
        dp1[0][0] = dp2[0][0] = 15000;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= m; j++) {
                if (j > i)    break;
                dp1[i][j] = 15001, dp2[i][j] = -15001;
                if (j < i) {
                    db a = (1 - 2.0*W[i]) * dp1[i - 1][j];
                    db b = (1 - 2.0*W[i]) * dp2[i - 1][j];
                    dp1[i][j] = min(a, b);
                    dp2[i][j] = max(a, b);
                }
                if (j) {
                    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];
                    dp1[i][j] = min(dp1[i][j], min(a, b));
                    dp2[i][j] = max(dp2[i][j], max(a, b));
                }
            }
        }
        db ans = 15001;
        for (int j = 0; j <= m; j++)    ans = min(ans, dp1[n][j]);
        printf("Case #%d: %.3lf
", kase, ans);
    }
    return 0;
}