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×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 xi 个)

∑xi=N

∑xi×i=2×N−2

求 max(f(i)×xi)

很容易想到一个 O(N3) 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(N2) 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;}