[hdu 5534]2015ACM/ICPC亚洲区长春站 Partial Tree 完全背包

Partial Tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 674 Accepted Submission(s): 331

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 are nn−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 is f(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

2
3
2 1
4
5 1 4

Sample Output

5
19

Source
2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5534

题意：
给一棵n个点的树添加边，给定度函数f(d)为一个点的度的函数，求所有点的度函数的最大和

思路：
树，所以每个点至少１个度；
然后dp作n-2个度；剩下n-2的度分给n个点
背包价值变成v[i]＝ｖ【ｉ】－ｖ【１】；
dp[i]代表i个度的最优值

 dp[i+j-1] = max(dp[i+j-1], dp[j] + F[i]);

代码：

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int T;
int ans;
int dp[2200];
int f[2200];
int n;
int solve(int m)
{
    for(int i=1;i<=n-2;i++) dp[i]=-99999999;
    dp[0]=0;
    for(int i=2;i<=n-1;i++)
    for (int j=0; j+i-1<=n-2;j++)
        dp[i+j-1]=max(dp[i+j-1],dp[j]+f[i]);

    return dp[n-2];
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        for(int i=1;i<n;i++) scanf("%d",&f[i]);
        ans=n*f[1];
        for(int i=2;i<n;i++) f[i]-=f[1];
        printf("%d\n",ans+solve(n-2));

    }   
}