2020 Multi-University Training Contest 2 Lead of Wisdom (爆搜，dfs）

题面

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

思路

爆搜题，直接dfs就好了。stl还是好用，哎，以后要学着多用stl，另外在比赛的时候要敢于去写代码，多找找感觉，会好的。

代码实现

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<vector>
#include<cmath>
#include<cstring>
using namespace std;
typedef long long ll;
struct node {
    int a,b,c,d;
};
long long ans;
int n,m;
vector <node > va[55];
void dfs (int num,int a,int b,int c,int d) {
    if (!num) {
        ans=max (ans,1ll*a*b*c*d);
        return;
    }
    int k=va[num].size ();
    if (!k) return dfs (num-1,a,b,c,d);
    for (int i=0;i<k;i++) dfs (num-1,a+va[num][i].a,b+va[num][i].b,c+va[num][i].c,d+va[num][i].d);
}

int main () {
   int t;
   cin>>t;
   while (t--) {
       cin>>n>>m;
       ans=0;
       for (int i=0;i<=m;i++) va[i].clear ();
       for (int i=1,ty,y,z,x,e;i<=n;i++) {
           cin>>ty>>x>>y>>z>>e;
           node t={x,y,z,e};
          va[ty].push_back (t);
       }
       dfs (m,100,100,100,100);
       printf ("%lld",ans);
   }
    

    return 0;
}