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-8;
const int mod = 10007;
const int maxn = 1e7 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;

short n, P, A, C, M;
short p[37], a[37], c[37], m[37], g[37], cnt[37];
short dp[37][37][37][37][37];
bool vis[37][37][37][37][37];

int main() {
    //FIN;
    cin >>n;
    for(int i = 1; i <= n; i++) {
        cin >>p[i] >>a[i] >> c[i] >>m[i] >>g[i];
    }
    cin >>P >>A >>C >>M;
    int flag1 = 0, flag2 = 0, flag3 = 0, flag4 = 0, mx = 0;
    for(int i = 1; i <= n; i++) {
        for(int j = 0; j <= P; j++) {
            for(int k = 0; k <= A; k++) {
                for(int l = 0; l <= C; l++) {
                    for(int t = 0; t <= M; t++) {
                        dp[i][j][k][l][t] = dp[i-1][j][k][l][t];
                        if(j >= p[i] && k >= a[i] && l >= c[i] && t >= m[i]) {
                            if(dp[i-1][j-p[i]][k-a[i]][l-c[i]][t-m[i]] + g[i] > dp[i-1][j][k][l][t]) {
                                dp[i][j][k][l][t] = dp[i-1][j-p[i]][k-a[i]][l-c[i]][t-m[i]] + g[i];
                                vis[i][j][k][l][t] = true;
                            }
                        }
                    }
                }
            }
        }
    }
    for(int j = 0; j <= P; j++) {
        for(int k = 0; k <= A; k++) {
            for(int l = 0; l <= C; l++) {
                for(int t = 0; 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 = 0;
    for(int i = n; i >= 1; i--) {
        if(vis[i][flag1][flag2][flag3][flag4]) {
            cnt[num++] = i - 1;
            flag1 -= p[i];
            flag2 -= a[i];
            flag3 -= c[i];
            flag4 -= m[i];
        }
    }
    cout <<num <<endl;
    for(int i = num - 1; i >= 0; i--) {
        cout <<cnt[i] <<" ";
    }
    cout <<"\n";
    return 0;
}

 

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+7;
const ll infll=0x3f3f3f3f3f3f3f3f;
const int MX = 1e5 + 7;
const int INF = 0x3f3f3f3f;
  
short n;
short dp[37][37][37][37];
bool vis[37][37][37][37][37];
short p[37],a[37],c[37],m[37],g[37];
short P,A,C,M;
  
int main(){
    //FIN;
    cin>>n;
    for(int i=1;i<=n;i++)
        cin>>p[i]>>a[i]>>c[i]>>m[i]>>g[i];
    cin>>P>>A>>C>>M;
    memset(dp,0,sizeof(dp));
    memset(vis,0,sizeof(vis));
    for(int i=1;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]=1;
                    }
                }
            }
        }
    }
    int ans[37],num=0;
    while(n){
        if(vis[n][P][A][C][M]){
            P-=p[n];
            A-=a[n];
            C-=c[n];
            M-=m[n];
            ans[++num]=n-1;
        }n--;
    }
    printf("%d\n",num);
    for(int i=num;i>=1;i--)
        printf("%d ",ans[i]);
    printf("\n");
    return 0;
}