多校第二场——1010 Lead of Wisdom

Problem Description
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 ai,bi,ci and di. 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+∑i∈Sai)(100+∑i∈Sbi)(100+∑i∈Sci)(100+∑i∈Sdi)

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 ti,ai,bi,ci and di (1≤ti≤k, 0≤ai,bi,ci,di≤100), denoting an item of type ti whose attributes are ai,bi,ci and di.

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

题意：有n个武器，都属于k种武器种其中一种。从n个武器种选出一个集合S，S种每种武器最多只能出现一次。每种武器有a,b,c,d四个值，均大于零。求DMG最大的S。
DMG=(100+∑i∈Sai)(100+∑i∈Sbi)(100+∑i∈Sci)(100+∑i∈Sdi)

思路：最开始以为又是动态规划，看到都头疼，后面看数据量和时间，用暴力dfs，居然过了。

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 60;
int cnt[N];
int next_[N];
int temp[N][N][4];
void dfs(int x, ll a, ll b, ll c, ll d, int m, ll &ans)
{
	if (x > m)
	{
		ll tmp = a * b*c*d;
		if (tmp > ans)
		{
			ans = tmp;
		}
		return;
	}
	int num = cnt[x];
	if (!num)
	{
		dfs(next_[x], a, b, c, d, m, ans);
		return;
	}
	for (int i = 1; i <= num; i++)
	{
		dfs(x + 1, a + temp[x][i][0], b + temp[x][i][1], c + temp[x][i][2], d + temp[x][i][3], m, ans);
	}
}
int main()
{
	int T;
	scanf("%d", &T);
	while (T--)
	{
		int n;
		int k;
		scanf("%d%d", &n, &k);
		for (int i = 1; i <= k; i++)
		{
			cnt[i] = 0;
		}
		int x;
		while (n--)
		{
			scanf("%d", &x);
			cnt[x]++;
			for (int j = 0; j < 4; j++)
			{
				scanf("%d", &temp[x][cnt[x]][j]);
			}
		}
		x = k + 1;
		for (int i = k; i; i--)
		{
			next_[i] = x;
			if (cnt[i])
			{
				x = i;
			}
		}
		ll ans = 0;
		dfs(1, 100, 100, 100, 100, k, ans);
		printf("%lld\n", ans);
	}
}