Partial Tree

http://acm.hdu.edu.cn/showproblem.php?pid=5534

Time Limit: / MS (Java/Others)    Memory Limit: / K (Java/Others)
Total Submission(s): Accepted Submission(s): 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− edges. You want to complete this tree by adding n− edges. There must be exactly one path between any two nodes after adding. As you know, there are nn− 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− integers f(),f(),…,f(n−). ≤T≤
≤n≤
≤f(i)≤
There are at most test cases with n>. Output
For each test case, please output the maximum coolness of the completed tree in one line. Sample Input Sample Output Source
2015ACM/ICPC亚洲区长春站-重现赛(感谢东北师大) Recommend
hujie

因为每个点最少的度数为1,所以可以先设每个点的度数为1,剩余的度数为(2*n-2)-n=n-2,相当于把剩下的n-2个度分给n个顶点,这样就转化为完全背包的题目了

 #include<bits/stdc++.h>
const int INF=0x3f3f3f3f;
using namespace std; int a[];
int dp[]; int main(){ int t;
scanf("%d",&t);
while(t--){
int n;
scanf("%d",&n);
for(int i=;i<n;i++){
scanf("%d",&a[i]);
}
int ans=a[]*n;
for(int i=;i<=n;i++) dp[i]=-INF;
dp[]=;
for(int i=;i<n;i++) a[i]-=a[];
for(int i=;i<n;i++){
for(int j=;j<=n-;j++){
dp[i+j-]=max(dp[i+j-],dp[j]+a[i]);
}
}
printf("%d\n",ans+dp[n-]);
} }

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