版权声明:本文为博主原创文章,未经博主允许不得转载. Rebuilding Roads Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 8227   Accepted: 3672 Description The cows have reconstructed Farmer John's farm, with its N barns (1 <= N <= 150, number 1..N) after the terrib…

Rebuilding Roads   Description The cows have reconstructed Farmer John's farm, with its N barns (1 <= N <= 150, number 1..N) after the terrible earthquake last May. The cows didn't have time to rebuild any extra roads, so now there is exactly one wa…

Rebuilding Roads Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 9105   Accepted: 4122 Description The cows have reconstructed Farmer John's farm, with its N barns (1 <= N <= 150, number 1..N) after the terrible earthquake last May. The…

Rebuilding Roads Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 10653 Accepted: 4884 Description The cows have reconstructed Farmer John's farm, with its N barns (1 <= N <= 150, number 1..N) after the terrible earthquake last May. The co…

Description The cows have reconstructed Farmer John's farm, with its N barns (1 <= N <= 150, number 1..N) after the terrible earthquake last May. The cows didn't have time to rebuild any extra roads, so now there is exactly one way to get from any g…

题目链接:http://poj.org/problem?id=1947 一共有n个节点,要求减去最少的边,行号剩下p个节点.问你去掉的最少边数. dp[u][j]表示u为子树根,且得到j个节点最少减去的边数. 考虑两种情况,去掉孩子节点v与去不掉. (1)去掉孩子节点:dp[u][j] = dp[u][j] + 1 (2)不去掉孩子节点:dp[u][j] = min(dp[u][j - k] + dp[v][k]) 综上就是dp[u][j] = min(dp[u][j] + 1, min(dp[…

题意: 有n个点组成一棵树,问至少要删除多少条边才能获得一棵有p个结点的子树? 思路: 设dp[i][k]为以i为根,生成节点数为k的子树,所需剪掉的边数. dp[i][1] = total(i.son) + 1,即剪掉与所有儿子(total(i.son))的边,还要剪掉与其父亲(+1)的边. dp[i][k] = min(dp[i][k],dp[i][j - k] + dp[i.son][k] - 2),即由i.son生成一个节点数为k的子树,再由i生成其他j-k个节点数的子树. 这里要还原i…

树形DP..... Rebuilding Roads Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 8188 Accepted: 3659 Description The cows have reconstructed Farmer John's farm, with its N barns (1 <= N <= 150, number 1..N) after the terrible earthquake last Ma…

Find Metal Mineral Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 3397    Accepted Submission(s): 1588 Problem Description Humans have discovered a kind of new metal mineral on Mars which are d…

题目大意:给一棵有根带点权树,并且给出容量.求在不超过容量下的最大权值.前提是选完父节点才能选子节点. 题目分析:树上的分组背包. ps:特判m为0时的情况. 代码如下: # include<iostream> # include<cstdio> # include<vector> # include<cstring> # include<algorithm> using namespace std; const int N=105; const…