1 01背包
求方案数
//01背包求方案数 AcWing 278 数字组合
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const int N = 110, M = 1e5 + 10;
int f[M], v[N];
int main()
{
int n, m;
cin >> n >> m;
for(int i = 1; i <= n; i ++ ) cin >> v[i];
f[0] = 1;// 总和为 0 有一种方案:一个都不选
for(int i = 1; i <= n; i ++ )
{
for(int j = m; j >= v[i]; j -- )
{
f[j] += f[j - v[i]];
}
}
cout << f[m] << endl;
return 0;
}
//完全背包 Acwing 1023买书
//每个物品有无限个
//求方案数
#include <bits/stdc++.h>
using namespace std;
const int N = 520, M = 6010;
int n, m;
int f[M];
int v[5] = {0, 10, 20, 50, 100};
int main()
{
cin >> m;
f[0] = 1;
for(int i = 1; i <= 4; i ++ )
{
for(int j = v[i]; j <= m; j ++ )
f[j] += f[j - v[i]];
}
cout << f[m] << endl;
return 0;
}
3多重背包
有n种物品 m为总容积 v[i]为价格, 体积 w[i]为价值 s为数量
朴素版
//多重背包 1019庆功会
// 朴素版 物品个数 * 总容量 * 物品数量 O(n * m * s)
#include <bits/stdc++.h>
using namespace std;
const int N = 520, M = 6010;
int n, m;
int f[M];
int main()
{
cin >> n >> m;
for(int i = 1; i <= n; i ++ )
{
int v, w, s;
cin >> v >> w >> s;
for(int j = m; j >= 0; j -- )
for(int k = 0; k <= s && k * v <= j; k ++ )
f[j] = max(f[j], f[j - k * v] + k * w);
}
cou << f[m] << endl;
return 0;
}
4分组背包
#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int n, m;
int v[N][N], w[N][N], s[N], f[N];
int main()
{
int n, m;
cin >> n >> m;
for(int i = 1; i <= n; i ++ )
{
cin >> s[i];
for(int j = 1; 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 = 1; k <= s[i]; k ++ )
if(j >= v[i][k]) f[j] = max(f[j], f[j - v[i][k]] + w[i][k]);
}
}
cout << f[m] << endl;
return 0;
}