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#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;
}
#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;
}