背包问题
01背包
状态f[i][j]
定义:前 i 个物品,背包容量 j 下的最优解(最大价值)
二维:状态转移方程dp[i][j] = max(dp[i-1][j], dp[i-1][j-v[i]] + w[i])
f[i-1][j]
:不选第 i 个物品的集合中的最大值
f[i-1][j-v[i]]+w[i]
:选第 i 个物品的集合,但是直接求不容易求所在集合的属性,先将第 i 个物品的体积减去,求剩下集合中选法的最大值
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
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 = 1; j <= m; j ++)
{
f[i][j] = f[i - 1][j];//如果不选,那么f[i][j]就等于上一个不包含i的f[i - 1][j]
if(j >= v[i]) f[i][j] = max(f[i][j], f[i - 1][j - v[i]] + w[i]);//包含i,就要选一下
}
cout << f[n][m] << endl;
return 0;
}
一维优化版
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
int n, m;
int v[N], w[N];
int f[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 = m; j >= v[i]; j --)
if(j >= v[i]) f[j] = max(f[j], f[j - v[i]] + w[i]);
//f[i] 仅用到了f[i-1]层, j与j-v[i] 均小于j,若用到上一层的状态时,从大到小枚举, 反之从小到大哦
cout << f[m] << endl;
return 0;
}
完全背包问题
朴素版
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
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 ++)
for(int k = 0; k * v[i] <= j; k ++)
f[i][j] = max(f[i][j], f[i - 1][j - k * v[i]] + w[i] * k);
cout << f[n][m] << endl;
}
二维优化版
dp[i][j - v] = max(dp[i - 1][j - v], dp[i - 1][j - 2 * v] + w, dp[i - 1][j - 3 * v] + 2 * w, .....)
;
而我们需要的dp[i][j]
的状态表示是:
dp[i][j]= max(dp[i - 1][j], dp[i - 1][j - v] + w, dp[i - 1][j - 2 * v] + 2 * w, dp[i - 1][j - 3 * v] + 3 * w)
;
将每一项一一比对,我们可以得到下列状态表示:
dp[i][j] = max(dp[i - 1][j], dp[i][j - v] +w)
;
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
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;
}
一维优化版
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1010;
int n, m;
int v[N], w[N];
int f[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 ++)
{
if(j >= v[i]) f[j] = max(f[j], f[j - v[i]] + w[i]);
}
cout << f[m] << endl;
return 0;
}
多重背包问题一
朴素版
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 110;
int n, m;
int v[N], w[N], s[N];
int f[N][N];
int main()
{
cin >> n >> m;
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 <= m; j ++)
for(int k = 0; k <= s[i] && k * v[i] <= j; k ++)
f[i][j] = max(f[i][j], f[i - 1][j - v[i] * k] + w[i] * k);
cout << f[n][m] << endl;
return 0;
}
多重背包问题二
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 25000, M = 2010;
int n, m;
int v[N], w[N];
int f[M];
int main()
{
cin >> n >> m;
int cnt = 0;//分的组
for(int i = 1; i <= n; i ++)
{
int a, b, s;
cin >> a >> b >> s;
int k = 1;//组里的个数
while(k <= s)
{
cnt ++;//组数 + 1
v[cnt] = a * k;//整体体积
w[cnt] = b * k;//整体价值
s -= k;
k *= 2;
}
if(s > 0)//剩下的
{
cnt ++;
v[cnt] = a * s;
w[cnt] = b * s;
}
}
n = cnt;
for(int i = 1; i <= n; i ++)
for(int j = m; j >= v[i]; j --)
f[j] = max(f[j], f[j - v[i]] + w[i]);
cout << f[m] << endl;
return 0;
}
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;
}