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A*,32MS,90道题目留念评估函数 f2(x)= p(x)+3s(x) p(x)是x结点和目标结点相比每个将牌“离家”的最短距离之和; s(x)是:每个将牌和目标相比,若该将牌的后继和目标中该将牌的后继不同, 则该将牌得2分,相同则该将牌得0分,中间位置有将牌得1分,没将牌得0分。 Followed by: Post your reply here: |
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