# DP

+++

### 一.背包问题

##### 1.01背包

``````#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;
}
``````

``````#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

``````#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;
}
``````

``````#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;
}
``````

``````#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;
}
``````

FUNDAMENTAL PART4 DP