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…

算法: dp[i][j]表示以i为根的子树要变成有j个节点的状态需要减掉的边数. 考虑状态转移的时候不考虑i的父亲节点,就当不存在.最后统计最少减去边数的 时候+1. 考虑一个节点时,有两种选择,要么剪掉跟子节点相连的边,则dp[i][j] = dp[i][j]+1; 要么不剪掉,则d[i][j] = max(dp[i][j], dp[i][k]+dp[son][j-k]); #include<cstdio> #include<cstring> #include<iostre…

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…

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…

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…

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[…

题目链接 题意 : 给你一棵树,问你至少断掉几条边能够得到有p个点的子树. 思路 : dp[i][j]代表的是以i为根的子树有j个节点.dp[u][i] = dp[u][j]+dp[son][i-j]-1,son是u的儿子节点.初始是将所有的儿子都断开,然后-1代表的是这个儿子我需要了,不断了. #include<cstdio> #include<cstring> #include<algorithm> #include<iostream> #define…

dp[x][y]表示以x为根的子树要变成有y个点..最少需要减去的边树... 最终ans=max(dp[i][P]+t)  < i=(1,n) , t = i是否为整棵树的根 > 更新的时候分为两种情况..一种是要从其这个孩子转移过来...枚举做01背包..更新出每个状态的最小值..或者说直接砍掉这个孩子..那么只需将所有的状态多加个砍边... 这里的枚举做01背包..意思是由于叶子节点要放多少进去不确定..叶子节点要放的大小以及本节点的空间都在枚举更新...这种概念就是泛化背包..本质上是0…