hdu 5534 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): 1451    Accepted Submission(s): 724

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

232 145 1 4

 

Sample Output

519

题意：给出n个点，要求你连n-1条边使得这个图成为一颗树，给出每个度数的权值，求总的权值和最大；

解：因为每个点一定要连一条边，所以先给每个点分配一度，总度数是2*（n-1）,还剩n-2度，所以是个完全背包，之所以可以用背包是因为，一定可以将f[1...n-1]的度数

分配出去；

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <map>#include <queue>#include <bits/stdc++.h>using namespace std;typedef long long LL;const int N = 3e5+7;int f[N], dp[3000];int main(){    int t;    scanf("%d", &t);    while(t--)    {        int n;        scanf("%d", &n);        for(int i=1;i<n;i++) scanf("%d", &f[i]);        int ans=(n*f[1]);        for(int i=2;i<n;i++) f[i]-=f[1];        for(int i=1;i<n-1;i++) f[i]=f[i+1];        memset(dp,-0x3f3f3f3f,sizeof(dp));        dp[0]=0;        for(int i=1;i<=n-2;i++)        {            for(int j=i;j<=n-2;j++)            {                dp[j]=max(dp[j],dp[j-i]+f[i]);            }        }        printf("%d\n",dp[n-2]+ans);    }    return 0;}