牛客网暑期ACM多校训练营（第三场） — A — PACM Team
题目：
链接：https://www.nowcoder.com/acm/contest/141/A
来源：牛客网时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
Special Judge, 64bit IO Format: %lld
题目描述
Eddy was a contestant participating in ACM ICPC contests. ACM is short for Algorithm, Coding, Math. Since in the ACM contest, the most important knowledge is about algorithm, followed by coding(implementation ability), then math. However, in the ACM ICPC World Finals 2018, Eddy failed to solve a physics equation, which pushed him away from a potential medal.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 pi, 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.
示例1
输入
2
1 0 2 1 10
1 0 2 1 21
1 0 2 1
输出
1
1
示例2
输入
1
2 1 1 0 31
1 0 2 1
输出
0
思路：
多维费用的背包问题，对于每件物品，具有pi,ai,ci,mi四种费用，具有gi的价值。选择这件物品必须同时付出四种费用。问怎么选择物品可以得到最大的价值。
状态转移方程：F[i,p,a,c,m] = max( F[i-1][p][a][c][m], F[i-1][p-Pi] [a-Ai][c-Ci][m-Mi] + Gi )
因为5维的数组可能超内存，所以可以采用逆序的循环优化掉第一维度（滚动数组）
在转移的过程要用一个bool型的数组记录一下，以便后面回溯找出背包加入了哪些物品
代码：
#include<cstdio> #include<algorithm> #include<vector> using namespace std; vector<int> ans; int maxn = 100; int p[40],a[40],c[40],m[40],g[40]; int P,A,C,M; int dp[40][40][40][40]; bool path[37][37][37][37][37]; int main() { int k; scanf("%d",&k); for(int i=1;i<=k;i++){ scanf("%d%d%d%d%d",&p[i],&a[i],&c[i],&m[i],&g[i]); } scanf("%d%d%d%d",&P,&A,&C,&M); for(int i=1;i<=k;i++){ for(int pp=P;pp>=p[i];pp--){ for(int aa=A;aa>=a[i];aa--){ for(int cc=C;cc>=c[i];cc--){ for(int mm=M;mm>=m[i];mm--){ int add = dp[pp-p[i]][aa-a[i]][cc-c[i]][mm-m[i]]+g[i]; if(add>dp[pp][aa][cc][mm]){ dp[pp][aa][cc][mm]=add; path[i][pp][aa][cc][mm]=1;///记录这里转移 } } } } } } ///回溯加入的物品 int pp = P,aa=A,cc=C,mm=M; for(int i=k;i>=1;i--){ if(path[i][pp][aa][cc][mm]){ ans.push_back(i-1); ///题目里物品的下标从0开始，但是代码里从1开始，所以减去1 pp-=p[i]; aa-=a[i]; cc-=c[i]; mm-=m[i]; } } printf("%d\n",ans.size()); for(int i=0;i<ans.size();i++){ if(i==ans.size()-1) printf("%d\n",ans[i]); else printf("%d ",ans[i]); } return 0; }