HDU 5534 Partial Tree （完全背包）

Partial Tree

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 262144/262144K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 0

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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 find a partial tree on the way home. This tree has n nodes but lacks of n−1 edges. You want to complete this tree by adding n−1 edges. There must be exactly one path between any two nodes after adding. As you know, there arenn−2 ways to complete this tree, and you want to make the completed tree as cool as possible. The coolness of a tree is the sum of coolness of its nodes. The coolness of a node isf(d), where f is a predefined function and d is the degree of this node. What's the maximum coolness of the completed tree?

Input

The first line contains an integer T indicating the total number of test cases.
Each test case starts with an integer n in one line,
then one line with n−1 integers f(1),f(2),…,f(n−1).

1≤T≤2015
2≤n≤2015
0≤f(i)≤10000
There are at most 10 test cases with n>100.

Output

For each test case, please output the maximum coolness of the completed tree in one line.

Sample Input

232 145 1 4

Sample Output

519

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大） 

题意：给定一棵树中每个度数所对应的权值以及树的结点数，求该树的最大权值。

思路：想到要构造一棵n结点的树，结点的总度数应为2n-2，其中肯定要有度数为1的结点，之后就不了了之了。本来想用贪心，但发现各个结点的度数其实是会相互牵制的。最后看了题解发现用完全背包很巧妙。首先令每个结点度数都为1，之后还差n-2度数，将n-2作为背包中的容量限制，每个结点增加的度数可以从1到n-2。

注意转移方程dp[i]=max(dp[i],dp[i-j]+v[j+1]-v[1])，其中dp[i]为凑够i度数的最大权值。最后结果为dp[n-2]+v[1]*n，因为之前初始化没有+v[1]，但是在动规的过程中每次减去了v[1]，所以最后要补回来。

#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>#include <math.h>using namespace std;#define INF 999999int v[2016];int dp[2016];int main(){        int t;    scanf("%d",&t);    while(t--){        int n;        scanf("%d",&n);        for(int i=1;i<n;i++)            scanf("%d",&v[i]);        for(int i=1;i<=n-2;i++){            dp[i]=-INF;        }        dp[0]=0;        int m=n-2;        for(int i=1;i<=m;i++){            for(int j=1;j<n-1;j++){                if(j<=i)                    dp[i]=max(dp[i],dp[i-j]+v[j+1]-v[1]);            }        }        printf("%d\n",dp[m]+n*v[1]);    }     return 0;}