2023.2.6 普及组 day2

[Yapink]

• 第一题：简单的01背包问题，可优化一维

 

#include <iostream>

using namespace std;

const int N =  1010;

int n, m;
int f[N];

int main()
{
	cin >> m >> n;
	for (int i = 1; i <= n; i ++ ) 
	{
		int v, w;
		cin >> v >> w;
		for (int j = m; j >= v; j -- )
			f[j] = max(f[j], f[j - v] + w);
	}
	
	cout << f[m] << endl;
	return 0;
}

• 第二题：简单的完全背包模型，需要注意记得开longlong，不然爆int。

 

#include <iostream>

using namespace std;

typedef long long LL;

const int N = 10000010;

int n, m;
LL f[N];

int main()
{
	scanf("%d%d", &m, &n);
	for (int i = 1; i <= n; i ++ ) 
	{
		int v, w;
		scanf("%d%d", &v, &w);
		for (int j = v; j <= m; j ++ ) 
			f[j] = max(f[j], f[j - v] + w);
	}
	
	printf("%lld\n", f[m]);
	return 0;
}

 第三题：dp，考虑从两个方向过来的方案，列出状态转移方程。

 

 

#include <iostream>

using namespace std;

const int N = 110;

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

int main()
{
    int T;
    cin >> T;
    while (T -- )
    {
        cin >> n >> m;
        for (int i = 1; i <= n; i ++ ) 
            for (int j = 1; j <= m; j ++ ) 
                cin >> w[i][j];
                
        for (int i = 1; i <= n; i ++ ) 
            for (int j = 1; j <= m; j ++ ) 
                f[i][j] = max(f[i - 1][j], f[i][j - 1]) + w[i][j];
        
        cout << f[n][m] << endl;
    }
    
    return 0;
}

第四题：多重背包的简化版，把每盆花的体积看成1，总体积是花盆数

 

#include <iostream>

using namespace std;

const int N = 110, mod = 1e6 + 7;

int n, m;
int a[N];
int f[N][N];
 
int main()
{
	cin >> n >> m;
	for (int i = 1; i <= n; i ++ ) cin >> a[i];
	
	f[0][0] = 1;
	for (int i = 1; i <= n; i ++ ) 
		for (int j = 0; j <= m; j ++ ) 
			for (int k = 0; k <= a[i]; k ++ ) 
				if (j >= k)
					f[i][j] = (f[i][j] + f[i - 1][j - k]) % mod;
	
	cout << f[n][m] << endl;
	return 0;		
}