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比赛题目链接 题意:有n个人每人拿着一把枪想要杀死n个怪兽,大写字母代表人,小写字母代表怪兽.A只能杀死a,B只能杀死b,如题目中的图所示,枪的弹道不能交叉.人和怪兽的编号分别是1到n,问是否存在能全部杀死的情况,如果存在则输出编号1到n的每个人杀死的怪兽的编号,如果不能输出"Impossible". 题解:贪心,用递归实现,判断相邻的是否能构成一对,优先构成相邻的,如果不能就递归到前面看是否能构成一对即可. #include<cstdio> #include<cst…
题意很容易转化到这样的问题:在一个强连通的有向图D中是否存在这样的集合划分S + T = D,从S到T集合的边权大于从T到S集合的边权. 即D(i, j)  > B(j, i) + D(j, i).或者等价地对任意集合划分:D(i, j) <= B(j, i) + D(j, i)(*). 实际上若存在可行流f,满足:D(i, j) <= f(i, j) <= B(i, j) + D(i, j),则有对于任意割满足式(*),即可以返回"happy". 关于可行流参…