一道相当有思维性的背包。。（为什么每次遇到背包总是死
首先考虑到一棵树的度数和为2n-2，所以问题就是把这2n-2个度数分配到n个点上去，而不必考虑树的形态什么的。。
然后相当于把一些体积为1..n-1的物品装进容量为n的背包里面，求只有n个物品的最大价值。。
由于既有n个物品的限制，又有n种物品，所以直接背包是O(n^3)
然后考验思维的地方就出现了。。只要将每一个物品的体积减1，就相当于把1..n-2物品装进容量为n-2的背吧里，而且此时已经没有必取n个物品的限制了，因为每个物品的体积已经至少是1。。
然后直接跑完全背包，复杂度O(n^2)
还是太菜了。。
#include<bits/stdc++.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge*j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define succ(x) (1<<(x))
#define lowbit(x) (x&(-x))
#define ll long long
#define eps 1e-8
#define mid (x+y>>1)
#define sqr(x) ((x)*(x))
#define NM 10005
#define nm 2000005
const double pi=acos(-1);
const int inf=1e9+7;
using namespace std;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return x*f;
}
int d[NM],a[NM],n;
int main(){
int _=read();while(_--){
n=read();inc(i,1,n-1)a[i]=read();mem(d);d[0]=a[1]*n;
inc(i,1,n-2)inc(j,i,n-2)d[j]=max(d[j],d[j-i]+a[i+1]-a[1]);
printf("%d\n",d[n-2]);
}
return 0;
}
Partial Tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2108 Accepted Submission(s): 1048
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）
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