算法提高课 -- 1

数字三角形模型

Acwing 1015.摘花生

对应闫式DP分析法可写出对应的代码：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 1010;
int T, n, m;
int f[N][N], w[N][N];
int main() {
    cin >> T;
    while (T--) {
        cin >> n >> m;
        memset(f, 0, sizeof f);
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                cin >> w[i][j];
            }
        }
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= m; j++) {
                // 下标从(1, 1)开始
                f[i][j] = max(f[i - 1][j], f[i][j - 1]) + w[i][j];
            }
        }
        cout << f[n][m] << endl;
    }
    return 0;
}

Acwing 1018.最低通行费

代码展示：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 1010;
int n;
int w[N][N], f[N][N];
int main() {
    cin >> n;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            cin >> w[i][j];
        }
    }
    memset(f, 0x3f, sizeof f);
    f[1][0] = 0; // 由于是从左上两个方向过来的，所以开始f[1][0] || f[0][1]谁是0都行
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            f[i][j] = min(f[i][j - 1], f[i - 1][j]) + w[i][j];
        }
    }
    cout << f[n][n] << endl;
    return 0;
}

Acwing 1027.方格取数

代码展示：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 15;
int n, x, y, w;
int g[N][N], f[N * 2][N][N];
int main() {
    cin >> n;
    while (cin >> x >> y >> w) g[x][y] = w;
    for (int s = 2; s <= n * 2; s++) {
        for (int i1 = 1; i1 <= n; i1++) {
            for (int i2 = 1; i2 <= n; i2++) {
                if (s - i1 <= n && s - i1 > 0 && s - i2 <= n && s - i2 > 0) {
                    int& v = f[s][i1][i2];
                    int x = g[i1][s - i1];
                    if (i1 != i2) x += g[i2][s - i2];
                    v = max(v, f[s - 1][i1 - 1][i2 - 1] + x);
                    v = max(v, f[s - 1][i1-  1][i2] + x);
                    v = max(v, f[s - 1][i1][i2 - 1] + x);
                    v = max(v, f[s - 1][i1][i2] + x);
                }
            }
        }
    }
    cout << f[2 * n][n][n] << endl;
    return 0;
}

Acwing 275.传纸条

对应的DP分析图就不再画了，因为和上一题一样的，无非就是输入方式换了一下。

代码展示：

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 55;
int n, m;
int g[N][N], f[N * 2][N][N];
int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cin >> g[i][j];
        }
    }
    for (int s = 2; s <= n + m; s++) {
        for (int i1 = 1; i1 <= n; i1++) {
            for (int i2 = 1; i2 <= n; i2++) {
                if (s - i1 <= m && s - i1 > 0 && s - i2 <= m && s - i2 > 0) {
                    int& v = f[s][i1][i2];
                    int x = g[i1][s - i1];
                    if (i1 != i2) x += g[i2][s - i2];
                    v = max(v, f[s - 1][i1 - 1][i2 - 1] + x);
                    v = max(v, f[s - 1][i1-  1][i2] + x);
                    v = max(v, f[s - 1][i1][i2 - 1] + x);
                    v = max(v, f[s - 1][i1][i2] + x);
                }
            }
        }
    }
    cout << f[n + m][n][n] << endl;
    return 0;
}