分析:a^b+2(a&b)=a+b  so->a^(-b)+2(a&(-b))=a-b 然后树状数组分类讨论即可 链接:http://www.ifrog.cc/acm/problem/1023 吐槽:这个题本来是mod(2^40),明显要用快速乘啊,但是用了以后狂T,不用反而过了,不懂出题人 #include <iostream> #include <algorithm> #include <cmath> #include <vector&g…

Magic boy Bi Luo with his excited tree Problem Description Bi Luo is a magic boy, he also has a migic tree, the tree has N nodes , in each node , there is a treasure, it's value is V[i], and for each edge, there is a cost C[i], which means every time…

题目链接: Magic boy Bi Luo with his excited tree Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1037    Accepted Submission(s): 298 Problem Description Bi Luo is a magic boy, he also has a migic…

// 树形DP CCPC网络赛 HDU5834 Magic boy Bi Luo with his excited tree // 题意:n个点的树,每个节点有权值为正,只能用一次,每条边有负权,可以走多次,问从每个点出发的最大获益 // 思路: // dp[i]: 从i点出发回到i点的最大值 // d[i][0] 从i点出发不回来的最大值 // d[i][1] 从i点出发取最大值的下一个点 // d[i][2] 从i点出发取次大值 // dfs1处理出这四个 // 然后,从1开始转移,分别DP…

Magic boy Bi Luo with his excited tree Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1058    Accepted Submission(s): 308 Problem Description Bi Luo is a magic boy, he also has a migic tree,…

题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=5834 Description Bi Luo is a magic boy, he also has a migic tree, the tree has N nodes , in each node , there is a treasure, it's value is V[i], and for each edge, there is a cost C[i], which means…

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题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5834 题目大意: 一棵N个点的有根树,每个节点有价值ci,每条树边有费用di,节点的值只能取一次,边权每次经过都要扣,问从每一个节点开始走最大能获得的价值. 题目思路: [树形动态规划] 首先用dfs求出从根1往下走的:节点u往下走最后回到节点u的最大值g[u],节点u往下走最后不回到u的最优值和次优值f[0][u],f[1][u] 接着考虑一个节点u,除了以上的情况还有可能是往它的父亲方向走,这…

树形dp. 先dfs一次处理子树上的最优解,记录一下回到这个点和不回到这个点的最优解. 然后从上到下可以推出所有答案.细节较多,很容易写错. #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector>…

http://acm.hdu.edu.cn/showproblem.php?pid=5834 题意: 一棵树上每个节点有一个价值$Vi$,每个节点只能获得一次,每走一次一条边要花费$Ci$,问从各个节点出发最多能收获多少价值. 思路: 需要考虑子节点和父亲节点两个方面.既然是这样,那就需要两次dfs来解决问题了,$dp[u][0]$表示从u出发最后还是回到u的最大价值和,$dp[u][1]$表示从u出发最后不回到u的最大价值和. 第一次dfs很显然就是计算出每个节点往其子树方向的$dp[u][0…