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范例代码中,为了计算s1(只有一个老师能授课的课程集合)和s2(至少有两个老师能授课的课程集合),作者引入了s0。但其实不需要s0。 对于雇佣第i名老师,下一轮次的dp改为: dp(i+1, s1^st[i], s2|(s1&st[i])即可。
dp(i+1, s1^st[i], s2|(s1&st[i])
The text was updated successfully, but these errors were encountered:
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范例代码中,为了计算s1(只有一个老师能授课的课程集合)和s2(至少有两个老师能授课的课程集合),作者引入了s0。但其实不需要s0。
对于雇佣第i名老师,下一轮次的dp改为:
dp(i+1, s1^st[i], s2|(s1&st[i])
即可。The text was updated successfully, but these errors were encountered: