Acwing 动态规划打卡

Posted 2021-05-14 一个经常掉线的人

Acwing 动态规划打卡

多重背包问题 I

https://www.acwing.com/problem/content/4/

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cstring>
#include<unordered_map>
using namespace std;
typedef long long LL;
int N,V;
int v[105],w[105],s[105];
int f[105][105];
int main(){
    cin>>N>>V;
    for(int i=1;i<=N;i++){
        cin>>v[i]>>w[i]>>s[i];
    }
    for(int i=1;i<=N;i++){
        for(int j=0;j<=V;j++){
            for(int k=0;k<=s[i];k++){
                if(j>=k*v[i]){
                    f[i][j]=max(f[i][j],f[i-1][j-k*v[i]]+k*w[i]);
                }
            }
        }
    }
    cout<<f[N][V];
  return 0;
}
//  freopen("testdata.in", "r", stdin);

多重背包问题 II

https://www.acwing.com/problem/content/5/
转化成01背包问题,同时划分背包时不一个个分,而是通过2的次方数拆分
这样也可以表示出1到n的所有数

#include <iostream>
#include<algorithm>
#include<cstdio>
#include<vector>
#include<queue>
#include<stack>
#include<cstring>
using namespace std;
int N,V;
struct Good{
    int v,w;
};
vector<Good> goods;
int v,w,s;
int dp[2005];
int main(){
    cin>>N>>V;
    while(N--){
        cin>>v>>w>>s;
        for(int k=1;k<=s;k*=2){
            s-=k;
            goods.push_back({k*v,k*w});
        }
        if(s>0){
            goods.push_back({s*v,s*w});
        }
    }
    for(auto g:goods){
        for(int j=V;j>=g.v;j--){
            dp[j]=max(dp[j],dp[j-g.v]+g.w);
        }
    }
    cout<<dp[V];
  return 0;
}

//  freopen("testdata.in", "r", stdin);

分组背包问题

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 110;

int n, m;
int v[N][N], w[N][N], s[N];
int f[N];

int main()
{
    cin >> n >> m;

    for (int i = 1; i <= n; i ++ )
    {
        cin >> s[i];
        for (int j = 0; j < s[i]; j ++ )
            cin >> v[i][j] >> w[i][j];
    }

    for (int i = 1; i <= n; i ++ )
        for (int j = m; j >= 0; j -- )
            for (int k = 0; k < s[i]; k ++ )
                if (v[i][k] <= j)
                    f[j] = max(f[j], f[j - v[i][k]] + w[i][k]);

    cout << f[m] << endl;

    return 0;
}