package class065;

// Floyd算法模版（洛谷）

// 测试链接 : https://www.luogu.com.cn/problem/P2910

// 请同学们务必参考如下代码中关于输入、输出的处理

// 这是输入输出处理效率很高的写法

// 提交以下所有代码，把主类名改成Main，可以直接通过

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.io.OutputStreamWriter;

import java.io.PrintWriter;

import java.io.StreamTokenizer;

public class Code02_Floyd {

public static int MAXN = 101;

public static int MAXM = 10001;

public static int[] path = new int[MAXM];

public static int[][] distance = new int[MAXN][MAXN];

public static int n, m, ans;

// 初始时设置任意两点之间的最短距离为无穷大，表示任何路不存在

public static void build() {

for (int i = 0; i < n; i++) {

for (int j = 0; j < n; j++) {

distance[i][j] = Integer.MAX_VALUE;

}

}

}

public static void main(String[] args) throws IOException {

BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

StreamTokenizer in = new StreamTokenizer(br);

PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

while (in.nextToken() != StreamTokenizer.TT_EOF) {

n = (int) in.nval;

in.nextToken();

m = (int) in.nval;

for (int i = 0; i < m; i++) {

in.nextToken();

path[i] = (int) in.nval - 1;

}

// 这道题给的图是邻接矩阵的形式

// 任意两点之间的边权都会给定

// 所以显得distance初始化不太必要

// 但是一般情况下，distance初始化一定要做

build();

for (int i = 0; i < n; i++) {

for (int j = 0; j < n; j++) {

in.nextToken();

distance[i][j] = (int) in.nval;

}

}

floyd();

ans = 0;

for (int i = 1; i < m; i++) {

ans += distance[path[i - 1]][path[i]];

}

out.println(ans);

}

out.flush();

out.close();

br.close();

}

public static void floyd() {

// O(N^3)的过程

// 枚举每个跳板

// 注意，跳板要最先枚举！跳板要最先枚举！跳板要最先枚举！

for (int bridge = 0; bridge < n; bridge++) { // 跳板

for (int i = 0; i < n; i++) {

for (int j = 0; j < n; j++) {

// i -> .....bridge .... -> j

// distance[i][j]能不能缩短

// distance[i][j] = min ( distance[i][j] , distance[i][bridge] + distance[bridge][j])

if (distance[i][bridge] != Integer.MAX_VALUE

&& distance[bridge][j] != Integer.MAX_VALUE

&& distance[i][j] > distance[i][bridge] + distance[bridge][j]) {

distance[i][j] = distance[i][bridge] + distance[bridge][j];

}

}

}

}

}

}