hdu5534 Partial Tree dp(难)

hdu5534 Partial Tree  dp(难)

Partial Tree

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 374    Accepted Submission(s): 209


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 n1 edges. You want to complete this tree by adding n1 edges. There must be exactly one path between any two nodes after adding. As you know, there are nn2 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 n1 integers f(1),f(2),,f(n1).

1T2015
2n2015
0f(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
将前n个度数分给n个点,剩下的就可以随便分了。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define MS0(a) memset(a,0,sizeof(a))

using namespace std;

typedef long long ll;
const int maxn=100100;
const int INF=(1<<29);

int f[maxn],n;
ll dp[maxn];

int main()
{
    int T;cin>>T;
    while(T--){
        cin>>n;
        REP(i,1,n-1) scanf("%d",&f[i]);
        ll ans=f[1]*n;
        MS0(dp);
        REP(i,0,n-2) dp[i]=-INF;
        dp[0]=0;
        REP(i,1,n-2){
            REP(j,0,i){
                dp[i]=max(dp[i],dp[i-j]+f[j+1]-f[1]);
                //cout<<i<<" "<<dp[i]<<endl;
            }
        }
        ans+=dp[n-2];
        printf("%I64d\n",ans);
    }
    return 0;
}
View Code

 ------可能是我小学生般的dp水平才觉得难,一定要狂补dp!

posted @ 2015-11-12 17:06  __560  阅读(267)  评论(0编辑  收藏  举报