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…

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…

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…

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…

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[…

题目大意 给定一棵n个结点的树,问最少需要删除多少条边使得某棵子树的结点个数为p 题解 很经典的树形DP~~~直接上方程吧 dp[u][j]=min(dp[u][j],dp[u][j-k]+dp[v][k]-1) 方程的意思是 以u结点为根保留j个结点需要删除的最少的边的条数,那么可以选择在某个子结点v中选择k个保留,其他结点保留j-k个,为什么需要-1呢,因为相当于把子树v衔接到结点u上,因此边u->v是不需要删除的,所以要-1 代码: #include <iostream> #inc…

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…