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p[n][i]=min(p[n-1][i+d],p[n][i-1]+pri[i])

Posted by luoyudong at 2010-03-23 21:08:29 on Problem 2465
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)

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