很显然是一个单源有向图最短路问题。鉴于 $N$ 的规模只有 $100$，且输入数据给出的是邻接矩阵，故考虑 Floyd。求出最短路后要注意输出的格式。时间复杂度 $O(N^3)$。

#include <cstdio>
#include <algorithm>
#include <utility>
#define INF 0x3f3f3f3f
using namespace std;
int f[105][105];
pair<int, int> p[105];
int main() {
	int n, m;
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= n; j++) {
			scanf("%d", &f[i][j]);
			if (!f[i][j] && i != j)
				f[i][j] = INF;
		}
	for (int k = 1; k <= n; k++)
		for (int i = 1; i <= n; i++)
			for (int j = 1; j <= n; j++)
				f[i][j] = min(f[i][j], f[i][k] + f[k][j]);
	for (int i = 1; i <= n; i++)
		p[i] = make_pair(f[m][i], i);
	sort(p + 1, p + n + 1);
	int lst = 0;
	printf("%d", m);
	for (int i = 2; i <= n; i++) {
		if (p[i].first == INF)
			break;
		if (lst < p[i].first) {
			while (lst < p[i].first)
				lst++, puts("");
			printf("%d", p[i].second);
		} else
			printf(" %d", p[i].second);
	}
	return 0;
}