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package class_2022_12_4_week;
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// Floyd算法解析
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// 本题测试链接 : https://www.luogu.com.cn/problem/P2910
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// 请同学们务必参考如下代码中关于输入、输出的处理
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// 这是输入输出处理效率很高的写法
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// 提交以下所有代码，把主类名改成Main，可以直接通过
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import java.io.BufferedReader;
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import java.io.IOException;
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import java.io.InputStreamReader;
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import java.io.OutputStreamWriter;
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import java.io.PrintWriter;
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import java.io.StreamTokenizer;
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public class Code02_ClearAndPresentDanger {
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public static int N = 100;
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public static int M = 10000;
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public static int[] path = new int[M];
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public static int[][] distance = new int[N][N];
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public static int n, m, ans;
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public static void main(String[] args) throws IOException {
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BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
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StreamTokenizer in = new StreamTokenizer(br);
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PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
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while (in.nextToken() != StreamTokenizer.TT_EOF) {
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n = (int) in.nval;
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in.nextToken();
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m = (int) in.nval;
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for (int i = 0; i < m; i++) {
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in.nextToken();
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path[i] = (int) in.nval - 1;
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}
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for (int i = 0; i < n; i++) {
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for (int j = 0; j < n; j++) {
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in.nextToken();
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distance[i][j] = (int) in.nval;
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}
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}
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floyd();
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ans = 0;
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for (int i = 1; i < m; i++) {
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ans += distance[path[i - 1]][path[i]];
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}
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out.println(ans);
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out.flush();
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}
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}
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public static void floyd() {
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// O(N^3)的过程
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// 枚举每个跳板
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// 注意! 跳板要最先枚举，然后是from和to
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for (int jump = 0; jump < n; jump++) { // 中途！
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for (int from = 0; from < n; from++) { // from
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for (int to = 0; to < n; to++) { // to
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if (distance[from][jump] != Integer.MAX_VALUE && distance[jump][to] != Integer.MAX_VALUE
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&& distance[from][to] > distance[from][jump] + distance[jump][to]) {
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distance[from][to] = distance[from][jump] + distance[jump][to];
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}
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}
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}
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}
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// 如下这么写是错的
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// for (int from = 0; from < n; from++) {
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// for (int to = 0; to < n; to++) {
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// for (int jump = 0; jump < n; jump++) {
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// if (distance[from][jump] != Integer.MAX_VALUE && distance[jump][to] != Integer.MAX_VALUE
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// && distance[from][to] > distance[from][jump] + distance[jump][to]) {
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// distance[from][to] = distance[from][jump] + distance[jump][to];
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// }
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// }
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// }
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// }
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}
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}
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