Problem Statement
There are N turkeys. We number them from 1 through N.
M men will visit here one by one. The i-th man to visit will take the following action:
- If both turkeys xi and yi are alive: selects one of them with equal probability, then eats it.
- If either turkey xi or yi is alive (but not both): eats the alive one.
- If neither turkey xi nor yi is alive: does nothing.
Find the number of pairs (i, j) (1≤i<j≤N) such that the following condition is held:
- The probability of both turkeys i and j being alive after all the men took actions, is greater than 0.
Constraints
- 2≤N≤400
- 1≤M≤105
- 1≤xi<yi≤N
我有个乱搞做法，和题解略微不同，感性上看很有道理，在这里mark一下，如果哪位能叉掉请在评论区告知我。
设bz[i][j]∈{0,1}表示i是否可能与j同时存活。
易于发现这个bz是有传递性的，即如果bz[x][y]=0,bz[y][z]=0，那么一定有bz[x][z]=0
每次的连边(x,y)相当于是令bz[x][y]=0，那么我们传递一次bz数组。
特殊地，如果在这次(x,y)之前，已经有bz[x][y]=0，那么这次以后，x和y必定都死了,die[x]=die[y]=1。
最后如果数对(i,j)满足die[x]=0 and die[y]=0 and bz[x][y]=1，就满足条件。
复杂度O(nm+n^2)
贴一下官方题解：