PACM Team(牛客第三场多校赛+dp+卡内存+打印路径）

题目链接（貌似未报名的不能进去）：https://www.nowcoder.com/acm/contest/141/A

题目：

题意：背包题意，并打印路径。

思路：正常背包思路，不过五维的dp很容易爆内存，比赛时无限爆，后面队友提醒用short就过了。不过也可以用滚动减少内存消耗，两种代码实现都贴一下吧~

五维代码实现如下：

 #include <set>
 #include <map>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <bitset>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstdlib>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 typedef long long ll;
 typedef unsigned long long ull;

 #define lson i<<1
 #define rson i<<1|1
 #define bug printf("*********\n");
 #define FIN freopen("D://code//in.txt", "r", stdin);
 #define debug(x) cout<<"["<<x<<"]" <<endl;
 #define IO ios::sync_with_stdio(false),cin.tie(0);

 const double eps = 1e-;
 const int mod = ;
 const int maxn = 1e7 + ;
 const double pi = acos(-);
 const int inf = 0x3f3f3f3f;
 const ll INF = 0x3f3f3f3f3f3f3f3f;

 short n, P, A, C, M;
 short p[], a[], c[], m[], g[], cnt[];
 short dp[][][][][];
 bool vis[][][][][];

 int main() {
     //FIN;
     cin >>n;
     for(int i = ; i <= n; i++) {
         cin >>p[i] >>a[i] >> c[i] >>m[i] >>g[i];
     }
     cin >>P >>A >>C >>M;
     int flag1 = , flag2 = , flag3 = , flag4 = , mx = ;
     for(int i = ; i <= n; i++) {
         for(int j = ; j <= P; j++) {
             for(int k = ; k <= A; k++) {
                 for(int l = ; l <= C; l++) {
                     for(int t = ; t <= M; t++) {
                         dp[i][j][k][l][t] = dp[i-][j][k][l][t];
                         if(j >= p[i] && k >= a[i] && l >= c[i] && t >= m[i]) {
                             if(dp[i-][j-p[i]][k-a[i]][l-c[i]][t-m[i]] + g[i] > dp[i-][j][k][l][t]) {
                                 dp[i][j][k][l][t] = dp[i-][j-p[i]][k-a[i]][l-c[i]][t-m[i]] + g[i];
                                 vis[i][j][k][l][t] = true;
                             }
                         }
                     }
                 }
             }
         }
     }
     for(int j = ; j <= P; j++) {
         for(int k = ; k <= A; k++) {
             for(int l = ; l <= C; l++) {
                 for(int t = ; t <= M; t++) {
                     if(dp[n][j][k][l][t] > mx) {
                         flag1 = j, flag2 = k, flag3 = l, flag4 = t;
                         mx = dp[n][j][k][l][t];

                     }
                 }
             }
         }
     }
     int num = ;
     for(int i = n; i >= ; i--) {
         if(vis[i][flag1][flag2][flag3][flag4]) {
             cnt[num++] = i - ;
             flag1 -= p[i];
             flag2 -= a[i];
             flag3 -= c[i];
             flag4 -= m[i];
         }
     }
     cout <<num <<endl;
     for(int i = num - ; i >= ; i--) {
         cout <<cnt[i] <<" ";
     }
     cout <<"\n";
     return ;
 }

滚动数组实现如下：

 #include <bits/stdc++.h>
 #define fi first
 #define se second
 #define lson l,m,rt<<1
 #define rson m+1,r,rt<<1|1
 #define lowbit(x) x&-x
 #define MP make_pair
 #define pb push_back
 #define debug(x) cout<<"x= "<<x<<endl;
 #define FIN freopen("in.txt","r",stdin);
 using namespace std;
 typedef long long ll;
 typedef vector<int> vi;
 typedef pair<int,int>pii;
 const int mod=1e9+;
 const ll infll=0x3f3f3f3f3f3f3f3f;
 const int MX = 1e5 + ;
 const int INF = 0x3f3f3f3f;

 short n;
 short dp[][][][];
 bool vis[][][][][];
 short p[],a[],c[],m[],g[];
 short P,A,C,M;

 int main(){
     //FIN;
     cin>>n;
     for(int i=;i<=n;i++)
         cin>>p[i]>>a[i]>>c[i]>>m[i]>>g[i];
     cin>>P>>A>>C>>M;
     memset(dp,,sizeof(dp));
     memset(vis,,sizeof(vis));
     for(int i=;i<=n;i++){
         for(int j=P;j>=p[i];j--){
             for(int k=A;k>=a[i];k--){
                 for(int l=C;l>=c[i];l--){
                     for(int o=M;o>=m[i];o--){
                         if(dp[j][k][l][o] > dp[j-p[i]][k-a[i]][l-c[i]][o-m[i]]+g[i]) continue;
                         dp[j][k][l][o]=dp[j-p[i]][k-a[i]][l-c[i]][o-m[i]]+g[i];
                         vis[i][j][k][l][o]=;
                     }
                 }
             }
         }
     }
     int ans[],num=;
     while(n){
         if(vis[n][P][A][C][M]){
             P-=p[n];
             A-=a[n];
             C-=c[n];
             M-=m[n];
             ans[++num]=n-;
         }n--;
     }
     printf("%d\n",num);
     for(int i=num;i>=;i--)
         printf("%d ",ans[i]);
     printf("\n");
     return ;
 }