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Re:Floyd不超才怪In Reply To:Floyd不超才怪 Posted by:T3 at 2005-08-15 17:50:12 for(k=1;k<=n;k++) for(i=1;i<=n;i++) for(j=1;j<=n;j++) a[i][j]|=a[i][k]&a[k][j]; floyd不会超吧.我的78ms,还有别的方法吗? Followed by:
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