Acwing------动态规划

动态规划

背包问题

状态表示

1.集合：所有只考虑前i个物品，且总体积不大于j的所有选法

2.属性：MAX

2.1 去掉k个物品i

2.2 求MAX，f【i - 1】【j - k * v】

2.3 再加回来k个物品i

状态计算：集合的划分

1. 0-1背包（Acwing-2）

朴素做法

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

一位数组优化

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

2.完全背包（Acwing-3）

朴素做法

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

优化做法

#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;
}

终极做法

#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-I（Acwing）

朴素做法

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

二进制优化

#include <iostream>
#include <algorithm>

using namespace std;

const int N = 25000, M = 2000;
int n, m, cnt = 0;
int v[N], w[N];
int f[N];

int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        int a, b, s, k = 1;
        cin >> a >> b >> s;
        while (k <= s) {
            cnt ++;
            v[cnt] = a * k;
            w[cnt] = b * k;
            s -= k;
            k *= 2;
        }
        if (s > 0) {
            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（Acwing-9）

#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;
}

线性DP

• 递推方程存在线性关系

数字三角形（Acwing-898）

#include <iostream>
using namespace std;

const int N = 510;

int n, INF = 1e9;
int a[N][N], f[N][N];

int main() {
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= i; ++j) {
            f[i][j] = -INF;
            cin >> 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], f[i - 1][j]) + a[i][j];
        }
    }
    
    int res = -INF;
    for (int i = 1; i <= n; ++i) res = max(res, f[n][i]);
    cout << res << endl;
    
    return 0;
}

最长上升子序列（Acwing-895）

#include <iostream>
using namespace std;

const int N = 1010, INF = 1e9;
int n, a[N], f[N];

int main() {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n; ++i) {
        f[i] = 1;
        for (int j = 1; j < i; ++j) {
            if (a[j] < a[i]) {
                f[i] = max(f[j] + 1, f[i]);
            }
        }
    }
    int res = 0;
    for (int i = 1; i <= n; ++i) res = max(res, f[i]);
    cout << res << endl;
    return 0;
}

最长上升子序列-II（Acwing-896）

#include <iostream>
using namespace std;

const int N = 100010;
int n, a[N], f[N];

int main() {
    cin >> n;
    for (int i = 0; i < n; ++i) cin >> a[i];
    
    int len = 0;
    f[0] = -2e9;
    for (int i = 0; i < n; ++i) {
        int l = 0, r = len;
        while (l < r) {
            int mid = l + r + 1 >> 1;
            if (f[mid] < a[i]) l = mid;
            else r = mid - 1;
        }
        len = max(len, r + 1);
        f[r + 1] = a[i];
    }
    
    cout << len << endl;
    return 0;
}

最长公共子序列（Acwing-897）

#include <iostream>
using namespace std;

const int N = 1010;
int n, m;
char a[N], b[N];
int f[N][N];

int main () {
    cin >> 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);
        }
    }
    
    cout << f[n][m] << endl;
    
    return 0;
}

最短编辑距离（Acwing-902）

#include <iostream>
#include <algorithm>
using namespace std;

const int N = 1010;
int n, m;
char a[N], b[N];
int f[N][N];

int main() {
    scanf("%d%s", &n, a + 1);
    scanf("%d%s", &m, b + 1);
    
    for (int i = 0; i <= m; ++i) f[0][i] = i;
    for (int i = 0; i <= n; ++i) f[i][0] = i;
    
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            f[i][j] = min(f[i - 1][j], f[i][j - 1]) + 1;
            if (a[i] == b[j]) f[i][j] = min(f[i][j], f[i - 1][j - 1]);
            else f[i][j] = min(f[i][j], f[i - 1][j - 1] + 1);
        }
    }
    cout << f[n][m] << endl;
    
    return 0;
}

区间DP

石子合并（Acwing-282）

#include <iostream>
#include <algorithm>
using namespace std;

const int N = 310;
int n, s[N], f[N][N];

int main() {
    cin >> n;
    for (int i = 1;i <= n; ++i) cin >> 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] = 1e9;
            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]);
            }
        }
    }
    cout << f[1][n] << endl;
    
    return 0;
}