图论 - 最短路 - Floyd

图论 - 最短路 - Floyd

题目链接： https://www.luogu.org/problem/P2910

代码：

#include <bits/stdc++.h>

using namespace std;

const int N = 110;

const int M = 10010;

int n, m;

int dis[N][N], ord[M];

void Floyd() {
	for (int k = 1; k <= n; k ++ ) {
		for (int i = 1; i <= n; i ++ ) {
			for (int j = 1; j <= n; j ++ ) {
				if (dis[i][j] > dis[i][k] + dis[k][j]) {
					dis[i][j] = dis[i][k] + dis[k][j];
				}
			}
		}
	}
}

int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= m; i ++ ) {
		scanf("%d", &ord[i]);
	}
	for (int i = 1; i <= n; i ++ ) {
		for (int j = 1; j <= n; j ++ ) {
			scanf("%d", &dis[i][j]);
		}
	}
	Floyd();
	int ans = 0;
	for (int i = 1; i < m; i ++ ) {
		ans += dis[ord[i]][ord[i + 1]];
	}
	printf("%d\n", ans);
	return 0;
}