FUNDAMENTAL PART4 DP

Posted 2020-12-20 scl0725

DP

+++

一.背包问题

1.01背包

二维数组状态转移

#include <iostream>

using namespace std;

const int N = 1010;

int v[N], w[N];
int n, m;
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 - 1][j - v[i]] + w[i]);
        }
        
    cout << f[n][m] << endl;
    
    return 0;
}

一维数组状态转移

#include <iostream>

using namespace std;

const int N = 1010;

int v[N], w[N];
int n, m;
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 -- )
            f[j] = max(f[j], f[j - v[i]] + w[i]);    
        
    cout << f[m] << endl;
    
    return 0;
}

2.完全背包问题

未优化做法（TLE）

#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]] + k * w[i]);
    
    cout << f[n][m] << endl;
    
    return 0;
}

优化做法1.

#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;
}

状态转移方程优化为一维数组（滚动数组）

#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 = v[i]; j <= m; j ++ )
            f[j] = max(f[j], f[j - v[i]] + w[i]);
    
    cout << f[m] << endl;
    
    return 0;
}

3.多重背包问题的二进制优化

#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;
    for (int i = 1; i <= n; i ++ )
    {
        int a, b, s;
        scanf("%d%d%d", &a, &b, &s);
        int k = 1;
        
        while(k <= s)
        {
            cnt ++ ;
            v[cnt] = a * k;
            w[cnt] = b * k;
            s -= k;
            k *= 2;
        }
        
        if(s > 0)
        {
            cnt ++ ;
            v[cnt] = s * a;
            w[cnt] = s * b;
        }
    }
    
    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;
}

5.分组背包问题

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 110;

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

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;
}

二.线性DP

1.数字三角形

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 510, INF = 0x3f3f3f3f;

int a[N][N], f[N][N];
int n;

int main()
{
    scanf("%d", &n);
    
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= i; j ++ )
            scanf("%d", &a[i][j]);
    
    for (int i = 0; i <= n; i ++ )
        for (int j = 0; j <= i + 1; j ++ )
            f[i][j] = -INF;
    
    f[1][1] = a[1][1];
    for (int i = 2; i <= n; i ++ )
        for (int j = 1; j <= i; j ++ )
            f[i][j] = max(f[i  -1][j - 1] + a[i][j], f[i - 1][j] + a[i][j]); 
    
    int res = -INF;
    for (int i = 1; i <= n; i ++ ) res = max(res, f[n][i]);
    
    printf("%d
", res);
    
    return 0;
}

2.最长上升子序列

1.O(n2)做法

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1010;

int a[N], f[N];
int n;

int main()
{
    scanf("%d", &n);
    
    for (int i = 1; i <= n; i ++ ) scanf("%d", a + i);
    
    for (int i = 1; i <= n; i ++ )
    {
        f[i] = 1;
        for (int j = 1; j < i; j ++ )
            if(a[i] > a[j])
                f[i] = max(f[i], f[j] + 1);
    }
    
    int res = 0;
    for (int i = 1; i <= n; i ++ ) res = max(res, f[i]);
    
    printf("%d
", res);
    
    return 0;
}

3.最长公共子序列

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 1010;

char a[N], b[N];
int f[N][N];
int n, m;

int main()
{
    scanf("%d%d", &n, &m);
    scanf("%s%s", a + 1, b + 1);
    
    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]);
            if(a[i] == b[j]) f[i][j] = max(f[i][j], f[i - 1][j - 1] + 1);
        }
        
    printf("%d
", f[n][m]);
    
    return 0;
}

4.石子合并（区间DP）

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 310;

int s[N];
int n;
int f[N][N];

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i ++ ) scanf("%d", s + i);
    
    for (int i = 1; i <= n; i ++ ) s[i] += s[i - 1];
    
    for (int len = 2; len <= n; len ++ )
        for (int i = 1; i + len - 1 <= n; i ++ )
        {
            int l = i, r = i + len - 1;
            f[l][r] = 1e8;
            
            for (int k = l; k < r; k ++ )
                f[l][r] = min(f[l][r], f[l][k] + f[k + 1][r] + s[r] - s[l - 1]);
        }
        
    printf("%d
", f[1][n]);
    
    return 0;
}