[Problem description]
In the game of Monster Hunter, players can acquire corresponding skills by embedding different decorative beads on equipment to improve their fighting ability.
It is known that there are 6 pieces of equipment on the hunter. Each piece of equipment may have several decorative holes. Each decorative hole has its own level. It can be inlaid with a decorative bead less than or equal to its own level (you can also choose not to inlay).
There are M types of decorative beads, numbered from 1 to M, corresponding to M types of skills. The i-th type of decorative beads has a level of Li and can only be embedded in decorative holes with a level greater than or equal to Li.
For the i-th skill, when the number of decorative beads equipped with the corresponding skill reaches Ki, the value of Wi(Ki) will be generated. The greater the number of similar skills embedded, the greater the value generated, that is, Wi(Ki − 1) <Wi(Ki). But each skill has an upper limit Pi (1 ≤ Pi ≤ 7). When the number of equipped beads exceeds Pi, only the value of Wi(Pi) will be generated.
For the given equipment and decorative bead data, solve how to inlay decorative beads to maximize the total value of 6 pieces of equipment.
[Input format]
Input lines 1 to 6, including descriptions of 6 pieces of equipment. Among them, the first integer Ni of the i represents the number of decorative holes of the i equipment. Ni integers are immediately followed, which respectively represent the grade L (1 ≤ L ≤ 4) of each decorative hole on the equipment.
Line 7 contains a positive integer M, which represents the number of decorative beads (skills).
Lines 8 to M + 7, each line describes the situation of a decorative bead (skill). The first two integers in each row, Lj (1 ≤ Lj ≤ 4) and Pj (1 ≤ Pi ≤ 7), respectively represent the grade and upper limit of the j-th decorative beads. Next, Pj integers, where the k-th number represents the value Wj(k) when the number of decorative beads in the equipment is k.
[Output format] The
output line contains an integer, indicating the maximum value that can be obtained.
【样例输入】
1 1
2 1 2
1 1
2 2 2
1 1
1 3
3
1 5 1 2 3 5 8
2 4 2 4 8 15
3 2 5 10
【样例输出】
20
answer
The C++ code is as follows
#include <iostream>
#include <cstring>
#include <aLgorithm>
using namespace std;
constexpr size_t MAXN = 55, MAXS = 305, MAXM = 1e4 + 5;
int n[MAXN], cnt[5], Lv[MAXM], p[MAXM], w[MAXM][MAXN];
int dp[MAXM][MAXS];
int m;
inline void Read() {
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
for (int i = 1; i <= 6; i++) {
cin >> n[i];
for (int j = 1, x; j <= n[i]; j++) {
cin >> x; cnt[x]++; }
}
cin >> m;
for (int i = 1; i <= m; i++) {
cin >> Lv[i] >> p[i];
for (int j = 1; j <= p[i]; j++)
cin >> w[i][j];
}
}
inline int DP() {
memset(dp, 0x80, sizeof dp), dp[0][0] = 0;
int ans = 0, sum = 0, tot = 0;
for (int L = 4; L >= 1; L--) {
sum += cnt[L];
for (int i = 1; i <= m; i++)
if (Lv[i] == L) {
tot++;
for (int k = 0; k <= sum; k++)
dp[tot][k] = dp[tot - 1][k];
for (int j = 1; j <= p[i]; j++)
for (int k = j; k <= sum; k++)
dp[tot][k] = max(dp[tot][k], dp[tot - 1][k - j] + w[i][j]);
}
}
for (int i = 0; i <= sum; i++) ans = max(ans, dp[tot][i]);
return ans;
}
signed main() {
Read();
cout << DP();
}
The java code is as follows
import java.util.*;
public class Main {
public static void main(String[] args) {
int []zhu,l= new int[]{
0,0,0,0,0,0};
int[][] zsk;
Scanner sc=new Scanner(System.in);
for(int i=0;i<6;i++) {
int ko=sc.nextInt();
for(int j=0;j<ko;j++) {
int li=sc.nextInt();
l[li+1]++;
}
}
int M=sc.nextInt();
int z=l[4]+l[5]+l[2]+l[3];
l[3]=l[2]+l[3];
l[4]=l[3]+l[4];
zsk=new int[M+1][z+1];
for(int i=0;i<z+1;i++)zsk[0][i]=0;
for(int i=0;i<M+1;i++)zsk[i][0]=0;
for(int i=1;i<M+1;i++) {
int bt=i;
int li=sc.nextInt();
int lim=sc.nextInt();
zhu=new int[lim];
for(int j=0;j<lim;j++)zhu[j]=sc.nextInt();
for(int i1=1;i1<l[li]+1;i1++)zsk[M][i1]=zsk[M-1][i1];
for(int i1=l[li]+1;i1<z+1;i1++) {
int max=0;
for(int j=0;i1-j-1>=l[li]&&j<zhu.length;j++)
max=Math.max(max, zhu[j]+zsk[i-1][i1-j-1]);
max=Math.max(max, zsk[i-1][i1]);
zsk[i][i1]=max;
}
}
System.out.println(zsk[M][z]);
}
}
The python solution is as follows
import os
import sys
s = [];r = [];m = []
for i in range(6):
s.append(list(map(int,input().split())))
n = int(input())
for i in range(n):
r.append(list(map(int,input().split())))
for i in range(6):
p = s[i]
for j in range(n):
if i == 0:
m.append(p[1:].count(int(j+1)))
else:
m[j] += p[1:].count(int(j+1))
'''print(m)''' '''用m[j]记录6件装备共有 j+1级孔m[j]个'''
mm = m'''假设j级孔全用j级珠子'''
up = [];down = [];'''up记录加一颗珠子提升的价值，down记录少一颗珠子减少的价值'''
def up1(k):
if mm[k]>=r[k][1]:
return 0
elif mm[k] == 0:
return r[k][2]
else:
return r[k][2+mm[k]]-r[k][1+mm[k]]
def down1(k):
if mm[k] == 1:
return r[k][2]
elif mm[k]>r[k][1]:
return 0
else :
return r[k][1+mm[k]]-r[k][mm[k]]
for i in range(n):
up.append(up1(i))
down.append(down1(i))
'''print(up,down)'''
'''第n+1级孔，如果1到n级孔中珠子加1的价值比它减少1颗的价值大，则数值各加1、减1，up、down更新'''
def main1(n):
if n == 0:
return
while max(up[:n])>down[n] and mm[n]>0:
t = up.index(max(up[:n]))
mm[t] += 1
up[t] = up1(t)
mm[n] -= 1
down[n] = down1(n)
main1(n-1)
return
point = 0
main1(n-1)'''输入有n级，为了和形参统一，此处减1'''
'''print(up,down)
print(mm)'''
for i in range(n):
if mm[i] != 0:
point += r[i][1+mm[i]]
print(point)'''得到总价值'''