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P[n][i]是到第n站，油为i时的最少费用，对f不是很明白，f = pri[n]？ 那方程是不是p[n][i]=min(p[n-1][i+d],p[n][i-1]+pri[i])In Reply To:就是p[n][i]=min(p[n-1][i+d],p[n][i-1])+f啊，d是前后两站的距离，f是钱，然后滚动数组就行了 Posted by:frkstyc at 2005-07-11 00:43:54 Followed by:
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