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二维dp

Posted by skogt at 2012-02-09 14:17:24 on Problem 3661
//dp[i][j]: 在第i分钟花费j的精力下所跑的最长的距离
//dp[i][j] = dp[i - 1][j - 1] + value[i];(j > 0)
//最终的输出dp[n][0]
//dp[i][0] = max(dp[i - 1][0], dp[i - j][j](i-j >= j));

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