Floyd弗洛伊德算法

B3611传递背包: https://www.luogu.com.cn/problem/B3611

视频讲解: https://www.bilibili.com/video/BV1iiksYCEgf/?spm_id_from=333.337.search-card.all.click

for (int k=1;k<=n;k++) {
	for (int i=1;i<=n;i++) {
		for (int j=1;j<=n;j++) {
			if (a[i][j] || (a[i][k] && a[k][j])) a[i][j] = 1; 
		}
	}
}

当i到j直接相连或从i到k再到j能到时，将a[i][j]设为可间接到达