2020杭电HDU-6768多校第二场Lead of Wisdom（暴力DFS）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=6772
博客园食用：https://www.cnblogs.com/lonely-wind-/p/13374448.html
CSDN食用：https://blog.csdn.net/qq_43906000/article/details/107569361

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:

$DMG=(100+{\sum }_{i\in S}{a}_i)(100+{\sum }_{i\in S}{b}_i)(100+{\sum }_{i\in S}{c}_i)(100+{\sum }_{i\in S}{d}_i)$

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 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

emmm，看题目。。。似乎可以爆搜诶，试了一发DFS。。。一发A了emmm，那就似乎没什么好说的了，不过我听说随机数也能过，榜上0ms估计都是随机数刚过去的吧QAQ，我也写了波随机数。。。一言难尽可能是我姿势不对吧。

以下是AC的DFS代码：

#include 
using namespace std;

typedef long long ll;
const int mac=100;

int tp[mac];
struct node
{
	int a,b,c,d;
};
vector<node>equip[mac];
ll ans=0;

void dfs(int a,int b,int c,int d,int num)
{
	if (num==0){
		ll sum=(100LL+a)*(100LL+b)*(100LL+c)*(100LL+d);
		ans=max(ans,sum);
		return;
	}
	int sz=equip[num].size();
	for (int i=0; i<sz; i++){
		a+=equip[num][i].a; b+=equip[num][i].b;
		c+=equip[num][i].c; d+=equip[num][i].d;
		dfs(a,b,c,d,num-1);
		a-=equip[num][i].a; b-=equip[num][i].b;
		c-=equip[num][i].c; d-=equip[num][i].d;
	}
}

int main(int argc, char const *argv[])
{
	int t;
	scanf ("%d",&t);
	while (t--){
		int n,k;
		scanf ("%d%d",&n,&k);
		int cnt=0;
		ans=0;
		for (int i=1; i<=k; i++) equip[i].clear(); 
		memset(tp,0,sizeof tp);
		for (int i=1; i<=n; i++){
			int type,a,b,c,d;
			scanf ("%d%d%d%d%d",&type,&a,&b,&c,&d);
			if (!tp[type]) equip[++cnt].push_back(node{a,b,c,d}),tp[type]=cnt;
			else equip[tp[type]].push_back(node{a,b,c,d});
		}
		dfs(0,0,0,0,cnt);
		printf("%lld\n",ans);
	}
	return 0;
}