动态规划——01背包

01背包问题==>每个数只能用一次

//二维
for (int i = 1; i <= n; i ++ )
	for (int j = 0; j <= m; j ++ )
	{
		dp[i][j] = dp[i-1][j];
		if(j >= v[i]) dp[i][j] = max(dp[i][j] , dp[i-1][j-v[i]] + w[i]);
	}

//一维
for (int i = 1; i <= n; i ++ )
	for (int j = m; j >= v[i]; j -- )
		dp[j] = max(dp[j],dp[j-v[i]]+w[i] );
		//

完全背包问题==>每个数可以用无限次

#include

using namespace std;

const int N = 1006;

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

int main()
{
	cin >> n >> m;
	for (int i = 1; i <= n; i ++ ) cin >> v[i] >> w[i];
	
	for (int i = 1; i <= n; i ++ )
		for (int j = 0; j <= m; j ++ )
		{
			f[i][j] = f[i-1][j];
			if(j >= v[i]) f[i][j] = max(f[i][j],f[i][j - v[i]] + w[i]);
		}
		
	cout << f[n][m] << endl;
	
	return 0;
 } 
//一维
for (int i = 1; i <= n; i ++ )
	for (int j = v[i]; j <= m; j ++ )
	{
		f[j] = max(f[j],f[j-v[i]] + w[i]);
		//