题目:这里 题意: 感觉并不能表达清楚题意,所以 Problem Description In mathematics, and more specifically in graph theory, a tree is an undirected graph in which any two nodes are connected by exactly one path. In other words, any connected graph without simple cycles is a…

Partial Tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 823    Accepted Submission(s): 407 Problem Description In mathematics, and more specifically in graph theory, a tree is an undirect…

Partial Tree Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5534 Description In mathematics, and more specifically in graph theory, a tree is an undirected graph in which any two nodes are connected by exactly…

题目链接: H - Partial Tree  HDU - 5534 题目大意:首先是T组测试样例,然后n个点,然后给你度数分别为(1~n-1)对应的不同的权值,然后问你在这些点形成树的前提下的所能形成的最大权值. 具体思路: 这个题是学长做的,我记录一下思路. 有点背包的感觉,但是和之前我做过的有点不同,原来的背包是互相不会影响的.但是对于这个题,我们需要的前提是形成一棵树,所以为了解决这个问题,我们先给每个点分配一个度,然后再去把剩余的n-2的度再分配下去就可以了. AC代码: #inclu…

一棵树一共有2*(n-1)度,现在的任务就是将这些度分配到n个节点,使这n个节点的权值和最大. 思路:因为这是一棵树,所以每个节点的度数都是大于1的,所以事先给每个节点分配一度,答案 ans=f[1]*n 先将答案赋值 所以接下来研究的就是,将剩下的n-2个度分配 即分别看 分配度数为1到n-2的节点的有几个(因为每个节点已经有一度),然后因为每个节点都加上了权值f[1],所以这时f[2]=f[2]-f[1],以此类推, 看到这里,就是一个完全背包问题:如果还没看出来,详细一点 有1到n-2这些…

Partial Tree In mathematics, and more specifically in graph theory, a tree is an undirected graph in which any two nodes are connected by exactly one path. In other words, any connected graph without simple cycles is a tree. You find a partial tree o…

Partial Tree Problem Description In mathematics, and more specifically in graph theory, a tree is an undirected graph in which any two nodes are connected by exactly one path. In other words, any connected graph without simple cycles is a tree. You f…

Partial Tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 228    Accepted Submission(s): 138 Problem Description In mathematics, and more specifically in graph theory, a tree is an undirect…

Partial Tree 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5534 完全背包 做这题前去学习了下完全背包,觉得这个优化简直神技!(以前都是用01背包做的,数据水的话可以过= =) ;i<=n;++i) for(int j=i;j<=V;++j) dp[j]=max(dp[j],dp[j-i]+v[i]); 完全背包 时间复杂度O(VE) 我们回到这道题,显然整棵树的总的度为2n-2,相当于将2n-2个度分配到n个结点中去: 为了保证…

Partial Tree http://acm.hdu.edu.cn/showproblem.php?pid=5534 Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): Accepted Submission(s): Problem Description In mathematics, and more specifically in graph theory, a tree…