hdu5534 Partial Tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 126 Accepted Submission(s): 68
Total Submission(s): 126 Accepted Submission(s): 68
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?
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.
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
注意到一个节点数为n的树的度数和玮2*n-2,所以问题就转换为了把2*n-2个度分配给n个节点所能获得的最大价值，而且每一个节点至少分到1个度。我们可以先每一个分一个度，然后把n-2个节点任意分配完。分配的时候因为已经分了1个度了，所以要把2~n-1的度看为1~n-1，然后做个完全背包就行了。
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 99999999
int v[2200],dp[2200];
int main()
{
int n,m,i,j,T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(i=1;i<=n-1;i++){
scanf("%d",&v[i]);
}
int ans=0;
ans+=v[1]*n;
for(i=2;i<=n-1;i++){
v[i]-=v[1];
}
for(i=1;i<=n-2;i++){
dp[i]=-inf;
}
dp[0]=0;
for(i=1;i<=n-2;i++){
v[i]=v[i+1];
}
for(i=1;i<=n-2;i++){
for(j=i;j<=n-2;j++){
dp[j]=max(dp[j],dp[j-i]+v[i]);
}
}
ans+=dp[n-2];
printf("%d\n",ans);
}
}
