【AGC016E】Poor Turkeys

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【AGC016E】Poor Turkeys

Description 有\(n\)(\(1 \le n \le 400\))只鸡,接下来按顺序进行\(m\)(\(1 \le m \le 10^5\))次操作.每次操作涉及两只鸡,如果都存在则随意拿走一只:如果只有一只存在,拿走这一只:如果都不存在,什么都不做. 求最后有多少对鸡(无序)可能共同存活. Solution 个人认为单用集合的解释方法有失偏颇. 首先考虑枚举两只鸡,规定它们一定要存活,然后模拟过程.怎么看单次模拟的复杂度都不会小于\(m\),因此要第一时间舍弃这种方法. 于是要换个…

[AGC016E]Poor Turkeys

[AGC016E]Poor Turkeys 题目大意: 有\(n(n\le400)\)只火鸡,编号为\(1\)到\(n\),有\(m(m\le10^5)\)个人,每人指定了两只火鸡\(x\)和\(y\). 若\(x\)和\(y\)都活着,那么这个人将会等概率地随机吃掉一只. 若\(x\)和\(y\)恰好活着一只,那么这个人将会吃掉活着的这只. 若\(x\)和\(y\)都已经死亡,那么只好什么都不做. 求有多少个\((i,j)(1\le i<j\le n)\)满足存在至少一种吃鸡方案使得在最终时刻…

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