图论+dp poj 1112 Team Them Up!

Floyd算法求任意两点间的最短距离+BFS+DFS 用邻接矩阵来存储图,Floyed算法求任意两点间的最短路径并输出,广度优先遍历,深度优先遍历 立即下载

题目链接:

http://poj.org/problem?id=1112

题目大意:

有编号为1~n的n个人,给出每个人认识的人的编号,注意A认识B,B不一定认识A,让你将所有的人分成两组,要求每组的人相互认识,且两组的人数要尽可能的接近。

求出每组的人的编号。

解题思路:

图论+背包(输出物品)。

相互认识的关系不好确定分组,如果转换思路,考虑相互不认识的情况就简单好多,如果A不认识B,且B不认识C,那么A和C必须分到同一组里。所以就想到了,连通分量的染色,相邻的两个染不同的颜色(0或1),每一个连通分量分成两组,并且相同颜色的人不能有边(一定要相互认识),容易知道不同连通分量的人一定相互认识,否则是连通的。

然后问题就转化为有多个连通分量,每个连通分量有两组,每组必须属于一个队,求两个队的人数差最小,并分别输出两队的人。

dp[i][j]表示到了第i个连通分量,且第一个队的人数为j时是否能够恰好凑齐。

动态规划、拔河比赛问题 题目描述 小明班里要举行一次拔河比赛,班主任决定将所有人分为两队,每个人都必须参加,两队人数之差不能超过1,并且两个队伍的体重之和要尽可能相近,当然相同是最好的了。 输入: 输入包含多组测试数据。 每组输入的第一行是一个正整数n(2<=n<=100),表示共有n个人。 接下来n行,每行输入一个整数w(1<=w<=450),表示每个人的体重。 输出 对于每组输入,分别输出两个队伍的体重之和,按升序排序。 立即下载

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