Lead of Wisdom
In an online game, "Lead of Wisdom" is a place where the lucky player can randomly get powerful items.
There are k types of items, a player can wear at most one item for each type. For the i
-th item, it has four attributes \(a_i,b_i,c_i\)
and \(d_i\). Assume the set of items that the player wearing is S
, the damage rate of the player DMG
can be calculated by the formula:
\(DMG=(100+\Sigma a_i)(100+\Sigma b_i)(100+\Sigma c_i)(100+\Sigma d_i)\)
Little Q has got n
items from "Lead of Wisdom", please write a program to help him select which items to wear such that the value of DMG is maximized.
Input
The first line of the input contains a single integer T
(1≤T≤10), the number of test cases.
For each case, the first line of the input contains two integers n
and k (1≤n,k≤50), denoting the number of items and the number of item types.
Each of the following n lines contains five integers \(t_i*,*a_i,b_i,c_i, d_i\)
$ (1≤t_i≤k,0≤a_i,b_i,c_i,d_i≤100)$, denoting an item of type \(t_i\)
whose attributes are $a_i,b_i,c_i $and \(d_i\)
.
Output
For each test case, output a single line containing an integer, the maximum value of DMG
.
Sample Input
1
6 4
1 17 25 10 0
2 0 0 25 14
4 17 0 21 0
1 5 22 0 10
2 0 16 20 0
4 37 0 0 0
Sample Output
297882000
暴力签到，对于这个数据规模直接爆搜即可，复杂度证明。可以用数组模拟一下链表，跳过没用的组。
当时没做出来，反思一下：当时写A有点怀疑人生，加上队友告诉我是个DP，加上判断题目类型的能力很弱....其实还是自己辣鸡T^T
这题本地跑了14s一度以为哪里出问题了，没想到交上去过了2333
#include <bits/stdc++.h>
using namespace std;
struct item
{
long long a, b, c, d;
};
vector<vector<item> > v;
int n, k;
long long ans = -1e18;
int Next[55];
void dfs(int x, long long a, long long b, long long c, long long d)
{
if(x == -1)
{
long long sum = a * b * c * d;
ans = max(ans, sum);
return;
}
for(int i = 0; i < v[x].size(); i++)
{
dfs(Next[x], a + v[x][i].a, b + v[x][i].b, c + v[x][i].c, d + v[x][i].d);
}
}
int main()
{
//freopen("1010.in","r",stdin);
//freopen("myout.out","w",stdout);
int Case;
cin >> Case;
for(int i = 1; i <= 51; i++)
{
vector<item> temp;
v.push_back(temp);
}
while(Case--)
{
cin >> n >> k;
ans = -1e18;
set<int>s;
//memset(Next, 0, sizeof(Next));
long long sa = 100, sb = 100, sc = 100, sd = 100;
for(int i = 1; i <= 51; i++)
{
v[i].clear();
}
for(int i = 1; i <= n; i++)
{
int t, a, b, c, d;
scanf("%d%d%d%d%d", &t, &a, &b, &c, &d);
v[t].push_back(item{a, b, c, d});
s.insert(t);
}
set<int>::iterator it;
int start, before;
bool began = 0;
for(it = s.begin(); it != s.end(); it++)
{
if(v[*it].size() == 1)
{
sa += v[*it][0].a;
sb += v[*it][0].b;
sc += v[*it][0].c;
sd += v[*it][0].d;
}
else if(!began)
{
start = before = *it;
began = 1;
}
else
{
Next[before] = *it;
before = *it;
}
}
Next[before] = -1;
dfs(start, sa, sb, sc, sd);
cout << ans << endl;
}
return 0;
}