[AcWing 1018] 最低通行费

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点击查看代码

#include<iostream>

using namespace std;

const int N = 110;
const int INF = 1e9;

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

int main()
{
	cin >> n;
	for (int i = 1; i <= n; i ++)
		for (int j = 1; j <= n; j ++)
			cin >> s[i][j];
	for (int i = 1; i <= n; i ++)
		for (int j = 1; j <= n; j ++) {
			if (i == 1 && j == 1)	f[i][j] = s[i][j];
			else if (i == 1)	f[i][j] = f[i][j - 1] + s[i][j];
			else if (j == 1)	f[i][j] = f[i - 1][j] + s[i][j];
			else f[i][j] = min(f[i - 1][j], f[i][j - 1]) + s[i][j];
		}
	cout << f[n][n] << endl;
	return 0;
}

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1. 状态表示
   \(f[i][j]\) 表示从 \((1,1)\) 到 \((i,j)\) 的所有路线的最小值
2. 状态计算
   ① \(f[1][1] = s[1][1]\)
   ② 第一行，\(i = 1\)，\(f[i][j] = f[i][j - 1] + s[i][j]\)
   ③ 第一列，\(j = 1\)，\(f[i][j] = f[i - 1][j] + s[i][j]\)
   ④ 其他情况，$f[i][j] = min(f[i - 1][j], f[i][j - 1]) + s[i][j] $