Acwing-----1018. 最低通行费

• 链接：https://www.acwing.com/problem/content/1020/

算法

• 状态表示：\(f(i, j)\)
  集合：所有从 \((1, 1)\) 到 \((i, j)\) 的路线
  属性：Min
• 状态计算：集合的划分
  \(f(i, j)\) 根据只能向东走和向南走分为 \(f(i-1, j)\) 和 \(f(i, j-1)\)
  需要进行特判是否为边界点

代码

#include <iostream>
using namespace std;

const int N = 110, INF = 1e9;
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];
        }
    }
    
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j) {
            if (i == 1 && j == 1) f[i][j] = w[i][j];
            else {
                f[i][j] = INF;
                if (i > 1) f[i][j] = min(f[i][j], f[i - 1][j] + w[i][j]);
                if (j > 1) f[i][j] = min(f[i][j], f[i][j - 1] + w[i][j]);
            }
        }
    }
    cout << f[n][n] << endl;
    return 0;
}