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一点思路首先把宽带b从小到大排序（离散化），并记录每个b的值是来自第几个设备的 假设我们已经找到了b的值为i的最小p的和，那么当我们将b的值更新到i+1时，只有存在b的值为i或b的值为i+1的设备的最小p需要更新，所以就只更新这些p，所以对于每一个设备，最多更新m^2*2次，所以总复杂度是n*m^2 Followed by: Post your reply here: |
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