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p[n][i]=min(p[n-1][i+d],p[n][i-1]+pri[i])In Reply To: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]) Posted by:sunmoonstar_love at 2005-07-11 01:06:25 牛逼的O（n^2），我的是O(n^3) Followed by: Post your reply here: |
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