链接：https://www.nowcoder.com/acm/contest/141/A
来源：牛客网
空间限制：C/C++ 262144K，其他语言524288K
Special Judge, 64bit IO Format: %lld
题目描述
Since then on, Eddy found that physics is actually the most important thing in the contest. Thus, he wants to form a team to guide the following contestants to conquer the PACM contests(PACM is short for Physics, Algorithm, Coding, Math).
There are N candidate groups each composed of pi physics experts, ai algorithm experts, ci coding experts, mi math experts. For each group, Eddy can either invite all of them or none of them. If i-th team is invited, they will bring gi knowledge points which is calculated by Eddy's magic formula. Eddy believes that the higher the total knowledge points is, the better a team could place in a contest. But, Eddy doesn't want too many experts in the same area in the invited groups. Thus, the number of invited physics experts should not exceed P, and A for algorithm experts, C for coding experts, M for math experts.
Eddy is still busy in studying Physics. You come to help him to figure out which groups should be invited such that they doesn't exceed the constraint and will bring the most knowledge points in total.
输入描述:
The first line contains a positive integer N indicating the number of candidate groups. Each of following N lines contains five space-separated integer p i, ai, ci, mi, gi indicating that i-th team consists of pi physics experts, ai algorithm experts, ci coding experts, mi math experts, and will bring gi knowledge points. The last line contains four space-separated integer P, A, C, M indicating the maximum possible number of physics experts, algorithm experts, coding experts, and math experts, respectively.
1 ≤ N ≤ 36 0 ≤ pi,ai,ci,mi,gi ≤ 36 0 ≤ P, A, C, M ≤ 36
输出描述:
The first line should contain a non-negative integer K indicating the number of invited groups. The second line should contain K space-separated integer indicating the index of invited groups(groups are indexed from 0).
You can output index in any order as long as each index appears at most once. If there are multiple way to reach the most total knowledge points, you can output any one of them. If none of the groups will be invited, you could either output one line or output a blank line in the second line.
case1：
输入
2
1 0 2 1 10
1 0 2 1 21
1 0 2 1
输出
1
1
case2：
输入
1 2 1 1 0 31
1 0 2 1
输出
0
题目大意：有一个四维背包，求取得最大值的方式（即有多少组，和具体组的 index）
思路：背包，但是要注意数据的类型要用short或者char，否则会爆内存。其次不能用滚动数组优化，因为要回溯到具体哪一组。做法一就是单纯的五维背包
状态：dp[ i ][ p ][ a ][ c ][ m ] 走了 i 组背包容量为 p, a, c, m 的最大值
状态转移： dp[ i ][ p ][ a ][ c ][ m ] = max(dp[i-1][ p ][ a ][ c ][ m ], dp[i-1][ p-ip[i] ][ a-ia[i] ][ c-ic[i] ][ m-im[i] ]);
这里回溯答案的话，用一个bool in[ i ][ p ][ a ][ c ][ m ]记录在当前状态有没有取第 i 组；
AC code：
#include <vector>
#include <cstdio>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
using namespace std; const int MAXN = ; short dp[MAXN][MAXN][MAXN][MAXN][MAXN];
int ip[MAXN], ia[MAXN], ic[MAXN], im[MAXN], g[MAXN];
bool in[MAXN][MAXN][MAXN][MAXN][MAXN];
int N, P, A, C, M; void slv()
{
for(int i = ; i <= N; i++) ///初始化
for(int p = ; p <= P; p++)
for(int a = ; a <= A; a++)
for(int c = ; c <= C; c++)
for(int m = ; m <= M; m++)
in[i][p][a][c][m] = false; for(int i = ; i <= N; i++)
{
for(int p = ; p <= P; p++) ///省去下面比较前一个取或者不取，一律初始化为不取
for(int a = ; a <= A; a++)
for(int c = ; c <= C; c++)
for(int m = ; m <= M; m++)
{
dp[i][p][a][c][m] = dp[i-][p][a][c][m];
}
for(int p = P; p >= ip[i]; p--)
for(int a = A; a >= ia[i]; a--)
for(int c = C; c >= ic[i]; c--)
for(int m = M; m >= im[i]; m--)
{
if(dp[i][p][a][c][m] < (dp[i-][p-ip[i]][a-ia[i]][c-ic[i]][m-im[i]]+g[i]))
{
dp[i][p][a][c][m] = dp[i-][p-ip[i]][a-ia[i]][c-ic[i]][m-im[i]]+g[i];
in[i][p][a][c][m] = true; ///取了第 i 组
}
}
}
vector<int>ans;
for(int i = N; i > ; i--)
{
if(in[i][P][A][C][M])
{
ans.push_back(i-); ///题目下标从 0 开始
P-=ip[i]; ///但是这里数组下标是从 1 开始
A-=ia[i];
C-=ic[i];
M-=im[i];
}
} printf("%d\n", (int)ans.size());
for(size_t i = ; i < ans.size(); i++)
{
printf("%d", ans[i]);
if(i+ < ans.size()) printf(" ");
else printf("\n");
}
}
int main()
{
scanf("%d", &N);
for(int i = ; i <= N; i++)
{
scanf("%d %d %d %d %d", &ip[i], &ia[i], &ic[i], &im[i], &g[i]);
}
scanf("%d %d %d %d", &P, &A, &C, &M);
slv();
return ;
}