杭电多校2020-2 Lead of Wisdom

http://acm.hdu.edu.cn/showproblem.php?pid=6772

题意：

n件物品，每种物品有一个种类ti
四个属性 ai , bi , ci , di
每个种类最多选一件物品，求 四个属性分别求和 再与100相加 最后 乘积

思路：

DFS 种类i物品数量不超过50 跳过没有的种类

（没过

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include <vector>
#include <iterator>
#include <utility>
#include <sstream>
#include <limits>
#include <numeric>
#include <functional>
using namespace std;
#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))
//#define sort(a,n,int) sort(a,a+n,less<int>())

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef char ch;
typedef double db;

const double PI=acos(-1.0);
const double eps=1e-6;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;
const int N=2e5+10;
const int mod=1e9+7;

int Len = 51;
struct item{
	int a = 0;
	int b = 0;
	int c = 0;
	int d = 0;
}s[N][N];

ll ans[N],nxt[N];
int main(){
	int sum = 0;
	int T = 0;
	int n = 0 , k = 0;
	cin >> T;
	while(T--){
		sum = 0;
        memset(ans , 0 , sizeof(ans));
        cin >> n >> k;
        for (int i = 0;i<n;i++){
            int t , q , w , e , r;
            cin >> t >> q >> w >> e >> r;
            s[t][ans[t]].a = q;
			s[t][ans[t]].b = w;
			s[t][ans[t]].c = e;
			s[t][ans[t]].d=r;
            ans[t]++;
        }
        int x=k+1;
        for(int i=k;i;i--){
            nxt[i]=x;
            if(ans[i])x=i;
        }

		int p = 1;
		int a , b , c , d;
		a = b = c = d = 0;
		int s1 = 0;
        for(int i = 0;i < ans[p] || ans[p] == 0;i++)
        {
    	 	if(k+1 == p)
    		{
        		ll s = (a+100)*(b+100)*(c+100)*(d+100);
        		if(s > s1)
				{
					s1 = s;
				}
        		return 0;
    		}
        	p += 1;
			a += s[p][i].a;
			b += s[p][i].b;
			c += s[p][i].c;
			d += s[p][i].d;
		}
        cout << sum;
	}
}