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
题意:老实说,开始看了半天并没有看懂题- -。
给你n个点,然后要求添加n-1条边,每个节点都有度(入度+出度),度的数量对应不同的权值,求树的最大权值
思路:总共有2*(n-1)个度,首先,每个点都先有一个度,然后分配剩下的n-2个度给n个点使它们最大,
于是就成了背包问题 /* 感觉动规怎么都不会,是时候去学学了
/*表示从题没看懂那一刻,就注定做不出来,于是只有去看别人的报告了- -
#include <iostream> #include <cstdio> #include <cstring> #include <cstdlib> #include <queue> #include <vector> #include <algorithm> #include <functional> //看懂题好不容易- - using namespace std; int a[20005]; int dp[20005]; int main() { int cas,n; scanf("%d",&cas); while(cas -- ) { scanf("%d",&n); for(int i = 0; i < n-1; i++) scanf("%d",a+i); int all = n-2; int ans = a[0]*n; for(int i = 1; i <= n; i++) dp[i] = -0x3f3f3f3f; dp[0]= ans; for(int i = 1; i < n-1; i++) a[i] -= a[0]; for(int i = 1; i <= all; i++) { for(int j = i; j <= all; j++) { dp[j] = max(dp[j],dp[j-i] + a[i]); } } printf("%d ",dp[all]); } return 0; }