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基于dp的floyd,简单for(k=1;k<=n;k++) for(i=1;i<=n;i++) for(j=1;j<=n;j++) if(g[i][k]>0&&g[k][j]>0) { int temp=max(g[i][k],g[k][j]); if(g[i][j]>0) g[i][j]=min(g[i][j],temp); else g[i][j]=temp; } Followed by: Post your reply here: |
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