hdu 5534 Partial Tree(dp+降唯)

题目链接

Partial Tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1577 Accepted Submission(s): 789

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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题意：

给出 $N$ 个结点我们需要添加 $N-1$ 条边使得它变成一棵树，定义一个函数 $f(x)$ ，给出它$f(1),f(2)...f(N-1)$ 的值，一个度为 $d$ 的结点的炫酷值是 $f(d)$，一棵树的炫酷值是所有结点的炫酷值之和，问 $N$ 个结点的树的最大炫酷值。

题解：

有个定理：如果对 $N$ 个结点指定其度，而且每个度都大于0且所有结点度之和为 $2 \times N-2$，那么能够构造出满足的一棵树。

那么问题就转化为(假设度为 $i$ 的点有 $x_i$ 个)

$\sum x_i=N$

$\sum x_i \times i=2 \times N-2$

求 $max(f(i) \times x_i)$

很容易想到一个 $O(N^3)$ 的 $dp$ ，令 $d[i][j]$ 表示分配了 $i$ 个点度数和为 $j$ 的最大炫酷值，那么答案就是 $d[N][N*2-2]$。可是会超时。

由于要给每个点都分配度数，且度数大于等于1，这一个限制使得 $dp$ 方程多了一维，因此我们可以先将每个结点分配度为1，每个 $f(i)-f(1)$，此时问题相当于一个完全背包，$O(N^2)$ 可解。

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define fi first
#define se second
#define dbg(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<endl;
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const int inf=0x3fffffff;
const ll mod=1000000007;
const int maxn=2017+10;
int d[maxn][maxn];
int v[maxn];
int main()
{
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        int n;
        scanf("%d",&n);
        rep(i,1,n) scanf("%d",&v[i]);
        rep(i,2,n) v[i]-=v[1];
        int ans=n*v[1];
        v[1]=0;
        rep(i,0,n) rep(j,0,n-1) d[i][j]=-1e9;
        d[0][0]=0;
        rep(i,1,n) rep(j,0,n-1)
        {
            d[i][j]=d[i-1][j];
            if(j>=i-1)
                d[i][j]=max(d[i][j],d[i][j-i+1]+v[i]);
        }
        ans+=d[n-1][n-2];
        printf("%d\n",ans);
    }
    return 0;
}

滚动数组：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define fi first
#define se second
#define dbg(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<endl;
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const int inf=0x3fffffff;
const ll mod=1000000007;
const int maxn=2017+10;
int d[maxn];
int v[maxn];
int main()
{
    int cas;
    scanf("%d",&cas);
    while(cas--)
    {
        int n;
        scanf("%d",&n);
        rep(i,1,n) scanf("%d",&v[i]);
        rep(i,2,n) v[i]-=v[1];
        int ans=n*v[1];
        v[1]=0;
        rep(j,0,n-1) d[j]=-1e9;
        d[0]=0;
        rep(i,1,n) rep(j,0,n-1)
        {
            //d[i][j]=d[i-1][j];
            if(j>=i-1)
                d[j]=max(d[j],d[j-i+1]+v[i]);
        }
        ans+=d[n-2];
        printf("%d\n",ans);
    }
    return 0;
}